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 N, ans;
unsigned long long A[1000000];
int BIT[1000000 + 1];
int datos[1000000 + 1];
void inserta(int i, int val) {
  while (i <= N) {
    BIT[i] += val;
    i += (i & -i);
  }
}
int query(int i) {
  int suma = 0;
  while (i > 0) {
    suma += BIT[i];
    i -= (i & -i);
  }
  return suma;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    int x = i - A[i];
    if (x < 0) x = 0;
    if (x != i) {
      inserta(x + 1, 1);
      inserta(i + 1, -1);
    }
  }
  for (int i = 0; i < N; i++) {
    if (query(i + 1) == 0) {
      ans++;
    }
  }
  cout << ans << "\n";
  return 0;
}
