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;
using ll = long long;
using ii = pair<int, int>;
using i3 = pair<int, ii>;
using li = pair<ll, int>;
using lii = pair<ll, ii>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<ii>;
using vpll = vector<pll>;
using vi3 = vector<i3>;
using vlii = vector<lii>;
const int N = 2e6 + 5;
const ll INF = 1e17 + 7;
const double eps = 1e-9, PI = acos(-1);
int n;
int ans[N];
void solve(int testCase) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    ans[max(0, i - x - 1)]++;
    ans[i - 1]--;
  }
  int res = 0;
  for (int i = n; i >= 1; i--) {
    ans[i] += ans[i + 1];
    if (ans[i] >= 0) res++;
  }
  printf("%d\n", res);
}
int main() {
  int t = 1;
  for (int testCase = 1; testCase <= t; testCase++) {
    solve(testCase);
  }
  return 0;
}
