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

3
1 1 1
3
1 1
2 1
3 1

2
1
1

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.

#include<bits/stdc++.h>

using namespace std;

const int MAXN=1e5+10;

const int INF=-0x3f3f3f3f;

int a[MAXN];

int dp[MAXN][510];

int main()

{

	int n;

	scanf("%d",&n);

    for (int i = 1; i <=n ; ++i) {

        scanf("%d",&a[i]);

    }

    for(int i=n;i>=1;i--) {//表示为p,由大->小，我们需要变化的次数增加

        for (int j = 1; j <=500; ++j) {//表示为k的大小

            if(i+a[i]+j>n) dp[i][j]=1;

            else

            dp[i][j]=dp[i+a[i]+j][j]+1;

        }

    }

    int ans=0;

    int m;

    scanf("%d",&m);

    int p,k;

    while(m--)

    {

        ans=0;

        scanf("%d%d",&p,&k);

        if(k>400)

        {

            while(p<=n)

            {

                p=p+a[p]+k;

                ans++;

            }

            printf("%d\n",ans);

        }

        else

            printf("%d\n",dp[p][k]);

    }

	return 0;

 }