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;
const double PI = 3.141592653589793238;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
int cnt, sum;
int arr[1000005];
int main() {
  int n, l, r;
  long long L;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &L);
    r = i;
    l = max(0ll, (long long)i - L);
    arr[l] += 1;
    arr[r] -= 1;
  }
  for (int i = 0; i < n; i++) {
    sum += arr[i];
    if (sum == 0) cnt++;
  }
  printf("%d\n", cnt);
  return 0;
}
