Wrath CodeForces

Hands that shed innocent blood!

There are n guilty people in a line, thei-th of them holds a claw with lengthL_i. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, thei-th person kills the j-th person if and only if j < i andj ≥ i - L_i.

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 integersL₁, L₂, ..., L_n (0 ≤ L_i ≤ 10⁹), whereL_i is the length of thei-th person's claw.

Output

Print one integer — the total number of alive people after the bell rings.

Example

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.

题意：

给定n个数据，两个数据之间相隔1米，给定的每个数，表示  所在位置能够击杀的最远的距离，方向值从右向左，

分析：

 最右边的那个人是肯定存活的，然后依次往右杀就行，求出最终活下来的人

到此为止，历经2天 终于来把代码补上了，

原因你们肯定不知道

先来科普一下：scanf是格式化输入 printf 是格式化输出 效率比较高，但是写代码麻烦，

cin输入流，cout是输出流,写起来虽然简单，但是效率稍低

AC：

#include <iostream>#include <cmath>#include <algorithm>#include <malloc.h>#include <math.h>#include <string.h>#include <cstdio>using namespace std;const  int inf=0x7fffffff; int n,a[1000010],ccmax=inf,sum=0;;int main(){   scanf("%d",&n);    for(int i=1;i<=n;i++)        scanf("%d",&a[i]);    for(int i=n;i>=1;i--)    {        if(i>=ccmax)            sum++;        ccmax=min(i-a[i],ccmax);    }    cout<<n-sum<<endl;}