Hands that shed innocent blood!

There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.

You are given lengths of the claws. You need to find the total number of alive people after the bell rings.

Input

The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.

Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.

Output

Print one integer — the total number of alive people after the bell rings.

Examples

Input

4
0 1 0 10


Output

1


Input

2
0 0


Output

2


Input

10
1 1 3 0 0 0 2 1 0 3


Output

3

Note

In first sample the last person kills everyone in front of him.
#include <bits/stdc++.h>
int arr[1000000][2];
int diff[1000001];
int max(int a, int b) { return (a > b) ? a : b; }
int main() {
  int i, n;
  scanf("%d", &n);
  int cnt = 0;
  for (i = 0; i < n; i++) {
    int ele;
    scanf("%d", &ele);
    if (i >= 1) {
      int low = max(i + 1 - ele, 1);
      if (low <= i) {
        arr[cnt][0] = low;
        arr[cnt][1] = i;
        cnt++;
      }
    }
  }
  int cnt1 = 0;
  for (i = 2; i <= n; i++) diff[i] = 0;
  for (i = 0; i < cnt; i++) {
    int low = arr[i][0], high = arr[i][1];
    if (low == 1) cnt1++;
    diff[low]++;
    if (high + 1 <= n) diff[high + 1]--;
  }
  int ans = 0;
  if (cnt1 == 0) ans++;
  int cur = cnt1;
  for (i = 2; i <= n; i++) {
    cur += diff[i];
    if (cur == 0) ans++;
  }
  printf("%d\n", ans);
  return 0;
}
