CF#804 B. Minimum number of steps（字符串，思维）

D. Minimum number of steps

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.

The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.

Input

The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.

Output

Print the minimum number of steps modulo 109 + 7.

Examples

input

ab

output

1

input

aab

output

3

Note

The first example: "ab"  →  "bba".

The second example: "aab"  →  "abba"  →  "bbaba"  →  "bbbbaa".

题意：将字符串中的 'ab' 改为 'bba' ，问至少改变多少次能使字符串不出现 'ab'

思路：我们的最终目的要将所有的 'a' 移到所有的 'b' 右边，而每移动一次，就会多产生一个 'b' ，所以我们只需计算每一个 'b'前面 'a'的个数 ，假设有 n 个 'a'，就需要移动 2^0 + 2^1 + …… +2^( n - 1) ，即 2^n - 1 次。

#include 
using namespace std;

const long long Mod = 1e9 + 7;
const long long N = 1e6 + 10;
char a[N];
long long b[N];
int n;

long long xx(long long a,long long n)
{
    long long ret = 1;
    long long tmp = a % Mod;
    while(n)
    {
        if(n & 1)
            ret = ret * tmp % Mod;
        tmp = tmp * tmp % Mod;
        n >>= 1;
    }
    return ret;
}
int main()
{
    while(~scanf("%s", a))
    {
        memset(b, 0, sizeof(b));
        int l = strlen(a);
        for(int i = 0; i < l; i++)
        {
            if(a[i] == 'a') b[i]++;
            if(i) b[i] += b[i - 1];
        }
        long long ans = 0;
        for(int i = 0; i < l; i++)
        {
            if(a[i] == 'b')
            {

                ans += xx(2, b[i]) - 1;
                ans %= Mod;

            }
        }
        printf("%lld\n", ans % Mod);
    }
}