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>
using namespace std;
const int mod = (int)10e9 + 7;
vector<int> readArr(int size);
template <class T>
void print1D(vector<T>);
template <class T>
void print2D(vector<vector<T>>);
const int N = (int)10e5 + 1;
vector<int> dp(N, 0);
void solve() {
  int n;
  cin >> n;
  vector<int> profit = readArr(n);
  dp[n - 1] = 1;
  for (int i = n - 1; i >= 0; --i) {
    if (profit[i + 1] >= profit[i])
      dp[i] = 1 + dp[i + 1];
    else
      dp[i] = 1;
  }
  int best = dp[0];
  for (int x : dp) best = max(best, x);
  cout << best;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt = 1;
  for (int i = 0; i < tt; i++) {
    solve();
  }
  return 0;
}
template <class T>
void print1D(vector<T> arr) {
  for (auto x : arr) {
    cout << x << " ";
  }
  cout << "\n";
}
template <class T>
void print2D(vector<vector<T>> arr) {
  for (auto x : arr) {
    for (auto el : x) {
      cout << el << " ";
    }
    cout << " \n";
  }
  cout << "\n";
}
vector<int> readArr(int size) {
  vector<int> arr(size, 0);
  for (int i = 0; i < size; i++) {
    cin >> arr[i];
  }
  return arr;
}
