Codeforces Round #446 (Div. 2) B. Wrath (贪心)

B. Wrath

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

40 1 0 10

output

1

input

20 0

output

2

input

101 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;int a[1000001];int main(){int n;cin >> n;for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);}int l = n - a[n], ans = 1;for(int i = n - 1; i >= 1; --i){if(i < l){ans++;}l = min(l, i - a[i]);}cout << ans << endl;}/*题意：1e6个人，每个人会杀死他前面a[i]个人，所有同时杀人，即即使一个会被杀死他也会杀人。问最多可以存活多少人。思路：可以发现最右边的人一定存活，然后从右往左贪心即可。*/