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 long long INF = (1LL << 60);
const double PI = 3.14159265359;
const double EPS = 1e-12;
const long long MOD = 1000000007;
template <typename T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <typename T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
long long dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
long long l[1000010], dp[1000010];
signed main(void) {
  long long n;
  scanf("%lld", &n);
  for (long long i = (long long)0; i < (long long)n; ++i) scanf("%lld", &l[i]);
  for (long long i = (long long)0; i < (long long)n; ++i) {
    dp[max(0LL, i - l[i])]++;
    dp[i]--;
  }
  for (long long i = (long long)1; i < (long long)n; ++i) {
    dp[i] += dp[i - 1];
  }
  long long cnt = 0;
  for (long long i = (long long)0; i < (long long)n; ++i) {
    if (dp[i] > 0) cnt++;
  }
  printf("%d\n", n - cnt);
  return 0;
}
