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 int mill = 1e9;
const int INF = 2e9;
const long long mod = 9999991;
const long long mod2 = 3e9 - 71;
const long long longINF = 3e18;
const long long ph = 31;
const long double pi = acos(-1);
const long double eps = 1e-6;
const int M = 1 << 21;
long long n, res;
int x;
vector<pair<int, int> > v;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (x != 0) {
      v.push_back({max(0, i - x), -1});
      v.push_back({i - 1, 1});
    }
  }
  sort(v.begin(), v.end());
  int cur = 0, c = 0;
  int k = (int)v.size();
  for (int i = 0; i <= n; i++) {
    while (c < k && v[c].first == i && v[c].second == -1) {
      cur++;
      c++;
    }
    if (cur == 0 && i != 0) res++;
    while (c < k && v[c].first == i && v[c].second == 1) {
      cur--;
      c++;
    }
  }
  cout << res;
  return 0;
}
