Codeforces 892 B.Wrath

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

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.

题目大意：n个人排队，第i个人会杀死第j个人当且仅当j < i并且j ≥i-li,问最后有多少个人能幸存下来.

分析：从后往前扫，这样就避免了j<i的干扰，记录一下i-li的最小值，就能判断第j个人是否会被杀死.

时间复杂度O(n).

#include <cstdio>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int inf = 0x7fffffff;

int n, l[], maxx = inf, ans;

int main()
{
    scanf("%d", &n);
    for (int i = ; i <= n; i++)
        scanf("%d", &l[i]);
    for (int i = n; i >= ; i--)
    {
        if (i >= maxx)
            ans++;
        maxx = min(maxx, i - l[i]);
    }
    printf("%d\n", n - ans);

    return ;
}