Codeforces Round #447 (Div. 2) A－C题解

作对一个题还涨分，我这分数是有多低，B题明知道爆表还不会改，难受。

A. QAQ

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.

Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).

Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.

Input

The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.

Output

Print a single integer — the number of subsequences "QAQ" in the string.

Examples

input

QAQAQYSYIOIWIN

output

4

input

QAQQQZZYNOIWIN

output

3

Note

In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".

大水题一发，100个字符，三层暴力都不超时，幸好这题手速A到，才加了50多分。

代码实现;

#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define mset(a,x) memset(a,x,sizeof(a))

using namespace std;
const double PI=acos(-1);
const int inf=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int mod=1e9+7;
int dir[4][2]={0,1,1,0,0,-1,-1,0};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}
ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}
char map[105];
int q[105],a[105];
int main()
{
	int n,i,j,k;
	while(~scanf("%s",map))
	{
		n=strlen(map);
		int l=0,r=0;
		int ans=0;
		for(i=0;i

B. Ralph And His Magic Field

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.

Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.

Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.

Input

The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).

Output

Print a single number denoting the answer modulo 1000000007.

Examples

input

1 1 -1

output

1

input

1 3 1

output

1

input

3 3 -1

output

16

Note

In the first example the only way is to put -1 into the only block.

In the second example the only way is to put 1 into every block.

找到规律容易吗？明知道10^18*10^18会爆，欧拉公式不会用啊，眼睁睁的看着这个题没过，难受啊。

题意是有n*m个格子的矩形，要求每个格子随便填数，使得结果每行每列和总的积为k，k只能是-1或1，问有多少种方法。

此题只能填1或-1，然后会发现当m+n是奇数的时候是无法构造的，即输出0. m*n的格子跟（m-1）*（n-1）的格子的摆放顺序有关，然后就是一波解释，不懂的可以百度，最后得到总数是2^(m-1)*(n-1)，解决掉爆表问题，就ok了。

代码实现：

#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define mset(a,x) memset(a,x,sizeof(a))

using namespace std;
const double PI=acos(-1);
const int inf=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int mod=1e9+7;
int dir[4][2]={0,1,1,0,0,-1,-1,0};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}
ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}

int main()
{
	ll a,b,x;
	while(cin>>a>>b>>x)
	{
		if(x==-1&&(a+b)%2)
		{
			cout<<"0"<

C. Marco and GCD Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In a dream Marco met an elderly man with a pair of black glasses. The man told him the key to immortality and then disappeared with the wind of time.

When he woke up, he only remembered that the key was a sequence of positive integers of some length n, but forgot the exact sequence. Let the elements of the sequence be a1, a2, ..., an. He remembered that he calculated gcd(ai, ai + 1, ..., aj) for every1 ≤ i ≤ j ≤ n and put it into a set S. gcd here means the greatest common divisor.

Note that even if a number is put into the set S twice or more, it only appears once in the set.

Now Marco gives you the set S and asks you to help him figure out the initial sequence. If there are many solutions, print any of them. It is also possible that there are no sequences that produce the set S, in this case print -1.

Input

The first line contains a single integer m (1 ≤ m ≤ 1000) — the size of the set S.

The second line contains m integers s1, s2, ..., sm (1 ≤ si ≤ 106) — the elements of the set S. It's guaranteed that the elements of the set are given in strictly increasing order, that means s1 < s2 < ... < sm.

Output

If there is no solution, print a single line containing -1.

Otherwise, in the first line print a single integer n denoting the length of the sequence, n should not exceed 4000.

In the second line print n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the sequence.

We can show that if a solution exists, then there is a solution with n not exceeding 4000 and ai not exceeding 106.

If there are multiple solutions, print any of them.

Examples

input

4
2 4 6 12

output

3
4 6 12

input

2
2 3

output

-1

Note

In the first example 2 = gcd(4, 6), the other elements from the set appear in the sequence, and we can show that there are no values different from 2, 4, 6 and 12 among gcd(ai, ai + 1, ..., aj) for every 1 ≤ i ≤ j ≤ n.

一道思维题，意思是找出一列数，使得任意两个数的gcd都在原数组中出现，任意数出这么一组数。

首先使新的序列构造出来，那么原序列中最小的数肯定是其他任何数的因子，构造的时候只需要把最小的数插入在原序列中间即可，这样可以保证gcd出来的结果不是自己就是最小的数。

代码实现：

#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define mset(a,x) memset(a,x,sizeof(a))

using namespace std;
const double PI=acos(-1);
const int inf=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int mod=1e9+7;
int dir[4][2]={0,1,1,0,0,-1,-1,0};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}
ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}
int map[maxn];

int main()
{
	int n,i,j,k;
	scanf("%d",&n);
	for(i=0;i