CF div2 321 A

A. Kefa and First Steps

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th day (1 ≤ i ≤ n) he makes a_i money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence a_i. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.

Help Kefa cope with this task!

Input

The first line contains integer n (1 ≤ n ≤ 10⁵).

The second line contains n integers a₁,  a₂,  ...,  a_n (1 ≤ a_i ≤ 10⁹).

Output

Print a single integer — the length of the maximum non-decreasing subsegment of sequence a.

Sample test(s)

Input

6
2 2 1 3 4 1

Output

3

Input

3
2 2 9

Output

3

Note

In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.

In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.

我居然上来就想套LIS的模板。。。我*了狗。。。然后发现读错题，扫一遍就行了。。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <vector>

using namespace std;

const int maxn = 1000005;

typedef long long LL;

vector<int>G[maxn];

int a[maxn];
int dp[maxn];


int main()
{
    int n,m;
    int tmp = 1;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    int cnt = 1;
    int ans = 0;
    while(cnt<=n){
        if(a[cnt]<=a[cnt+1]){
            tmp++;
        }
       else if (a[cnt]>a[cnt+1]){
            tmp  = 1;
        }
        ans = max(tmp,ans);
        cnt++;
    }




    printf("%d
",ans);


    return 0;
}