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 = 1e15 + 15;
int main() {
  cin.tie(0);
  cout.tie(0);
  cin.sync_with_stdio(false);
  long long n, temp;
  vector<long long> v;
  cin >> n;
  while (n--) {
    cin >> temp;
    v.push_back(temp);
  }
  vector<long long> x(v.size(), -1);
  for (long long i = v.size() - 1; i >= 0; i--) {
    long long kill = i - v[i];
    kill = max(kill, 0LL);
    if (i && kill != i) x[kill] = max(x[kill], i);
  }
  set<long long> ms;
  long long killed = 0;
  for (long long i = 0; i < x.size(); i++) {
    if (x[i] != -1) {
      ms.insert(x[i]);
    }
    if (ms.find(i) != ms.end()) ms.erase(ms.find(i));
    if (ms.size()) killed++;
  }
  cout << x.size() - killed;
  return 0;
}
