Codeforces Round #333 (Div. 2) B. Approximating a Constant Range

B. Approximating a Constant Range

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?

You're given a sequence of n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.

A range [l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.

Find the length of the longest almost constant range.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Sample test(s)

input

5
1 2 3 3 2

output

4

input

11
5 4 5 5 6 7 8 8 8 7 6

output

5

Note

In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].

题目大意：

给你一串数，让你求最大的区间（最大值-最小值<=1）

解题思路：

我是用集合做的。。。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define MM(a) memset(a,0,sizeof(a))

typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e5+5;
const int mod = 1e9+7;
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
LL gcd(LL a, LL b)
{
    if(b == 0)
        return a;
    return gcd(b, a%b);
}
multiset  s;
int arr[maxn];
int main()
{
    int n;
    while(cin>>n)
    {
        s.clear();
        for(int i=0; i 1)
            {
                s.erase(s.find(arr[i]));
                i++;
            }
            ret = max(ret, int(s.size()));
            j++;
        }
        cout<