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 ai money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence ai. 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, ..., an (1 ≤ ai ≤ 109).

Output

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

Examples

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.
#include <bits/stdc++.h>
const int MX = 1000 * 1000 + 24;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long double dist(long double x, long double y, long double a, long double b) {
  return sqrt((x - a) * (x - a) + (y - b) * (y - b));
}
using namespace std;
int n, tab[MX], v[MX];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int ans = 0;
  cin >> n;
  for (long long K = 0; K < n; K++) {
    cin >> tab[K];
  };
  for (int k = 0; k < n; k++) {
    v[k] = 1;
    if (tab[k - 1] <= tab[k]) v[k] = max(v[k], v[k - 1] + 1);
    ans = max(ans, v[k]);
  }
  cout << ans;
  return 0;
}
