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>
using namespace std;
int a[1500000];
int n;
int b[1500000];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int sum = 0;
  for (int i = n; i >= 1; --i) {
    if (a[i] > 0) {
      if (i - a[i] < 0) {
        for (int j = i - 1; j >= 1; --j) b[j] = 1;
        break;
      } else {
        int x = i - a[i];
        if (b[x] == 1)
          continue;
        else {
          for (int j = x; j <= i - 1; j++) {
            if (b[j] != 1)
              b[j] = 1;
            else
              break;
          }
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (b[i] == 0) {
      sum++;
    }
  }
  printf("%d\n", sum);
  return 0;
}
