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;
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const int N = (int)(4e6 + 5);
bitset<N> st;
inline int left(int p) { return p << 1; }
inline int right(int p) { return (p << 1) + 1; }
inline int mid(int L, int R) { return (L + R) / 2; }
void update(int p, int L, int R, int i, int j) {
  if (i > R || j < L) return;
  if (L >= i && R <= j) {
    st[p] = 0;
    return;
  }
  update(left(p), L, mid(L, R), i, j);
  update(right(p), mid(L, R) + 1, R, i, j);
}
int query(int p, int L, int R, int i, bool flag) {
  if (i > R || i < L) return -1;
  if (L == R)
    return (st[p] && flag);
  else {
    int p1 = query(left(p), L, mid(L, R), i, st[p] && flag);
    int p2 = query(right(p), mid(L, R) + 1, R, i, st[p] && flag);
    return (p1 != -1) ? p1 : p2;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<long long> v(n + 1);
  for (int(i) = (1); (i) <= (n); (i)++) cin >> v[i];
  st.set();
  for (int(i) = (1); (i) <= (n); (i)++) {
    int R = i - 1;
    int L = i - v[i];
    if (L < 1) L = 1;
    if (v[i] != 0) update(1, 1, n, L, R);
  }
  int ans = 0;
  for (int(i) = (1); (i) <= (n); (i)++) ans += query(1, 1, n, i, 1);
  cout << ans << '\n';
  return 0;
}
