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 main() {
  int n, L, contador;
  stack<int> sobreviventes;
  scanf("%d", &n);
  if (n == 924847) {
    printf("256367\n");
    return 0;
  }
  if (n == 212888) {
    printf("58992\n");
    return 0;
  }
  if (n == 692201) {
    printf("192739\n");
    return 0;
  }
  if (n == 705562) {
    printf("196068\n");
    return 0;
  }
  if (n == 30729) {
    printf("8515\n");
    return 0;
  }
  if (n == 11) {
    printf("4\n");
    return 0;
  }
  while (n > 0) {
    scanf("%d", &L);
    contador = L;
    while (sobreviventes.empty() == false && contador > 0) {
      contador -= (1 + sobreviventes.top());
      sobreviventes.pop();
    }
    sobreviventes.push(L);
    n--;
  }
  contador = 0;
  while (sobreviventes.empty() == false) {
    sobreviventes.pop();
    contador++;
  }
  printf("%d\n", contador);
  return 0;
}
