E. Array Queries

E. Array Queries

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

a is an array of n positive integers, all of which are not greater than n.

You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.

Input

The first line contains one integer n (1 ≤ n ≤ 100000).

The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n).

The third line containts one integer q (1 ≤ q ≤ 100000).

Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).

Output

Print q integers, ith integer must be equal to the answer to ith query.

Example

input

31 1 131 12 13 1

output

211

Note

Consider first example:

In first query after first operation p = 3, after second operation p = 5.

In next two queries p is greater than n after the first operation.

根据题意写就行了，主要是纯暴力会TLE，所以需要dp一下，大概找个区域，这里取350，大于根号100000就行。然后打个表

代码如下：

////  main.cpp//  E. Array Queries////  Created by 徐智豪 on 2017/4/16.//  Copyright © 2017年 徐智豪. All rights reserved.//#include <iostream>using namespace std;int n;int a[100005]={0};int q,p,k;int dp[100005][350]={0};int main(int argc, const char * argv[]) {    cin>>n;    for(int i=1;i<=n;i++)        cin>>a[i];    cin>>q;    for(int i=n;i>=1;i--)        for(int j=1;j<350;j++)        {            if(i+a[i]+j>n)dp[i][j]=1;            else            {                dp[i][j]=dp[i+a[i]+j][j]+1;            }        }    for(int j=1;j<=q;j++)    {        cin>>p>>k;        if(k<350)        {            cout<<dp[p][k]<<endl;        }        else        {            int count=1;            while((p=p+a[p]+k)<=n)                count++;            cout<<count<<endl;        }    }    return 0;}