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.

Example

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.

题意：

听说穿女装打acm有buff加成，但是死神达达讨厌女装，所以他要让n名女装大佬自相残杀。 一行中有n名女装大佬，每名女装大佬手中都拿着Li长度的镰刀。当12点钟声响起，每个人都会杀死在他左边的Li个人，该项动作同时进行。死神达达想知道最后存活下来的人数。

题解：

从后往前模拟即可，看代码即可。

#include
    
     #include
     
      using namespace std;const int maxn=1e6+5;int a[maxn];int main(){    int n;    while(cin>>n)    {        for(int i=0;i
      
       >a[i];        int ans=0;        int k=0;        for(int i=n-1;i>=0;i--)        {            if(k==0)                ans++;            k=max(a[i],k-1);        }        cout<
       
        <