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;
template <typename t>
void mi(t& a, const t& b) {
  a = min(a, b);
}
template <typename t>
void ma(t& a, const t& b) {
  a = max(a, b);
}
constexpr int g1 = 1000 * 1000 * 1000, si = 1024 * 1024, si_log = 10 + 10,
              mod = static_cast<int>(1e9) + 7;
constexpr long double eps = 1.0e-9;
int n, ar[si], ra[si];
char car[si];
vector<int> gr[si];
int64_t sol() {
  auto r = 0 * numeric_limits<int64_t>::max();
  for (int i = 1; i <= n; ++i) {
    const auto le = max(0, i - ar[i]), ri = i - 1;
    if (le <= ri) {
      ma(ra[le], ri);
    }
  }
  for (int i = 0; i <= n;) {
    if (0 == ra[i]) {
      ++i;
      continue;
    }
    auto mv = ra[i];
    while (i <= mv) {
      car[i] = 0;
      ++i;
      ma(mv, ra[i]);
    }
  }
  for (int i = 1; i <= n; ++i) r += car[i];
  return r;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  while (1 == scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) {
      scanf("%d", ar + i);
      car[i] = 1;
    }
    auto s = sol();
    cout << s << '\n';
    break;
  }
  return 0;
}
