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

3
1 1 1
3
1 1
2 1
3 1


Output

2
1
1

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.
import java.util.*;
import java.io.*; 
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    void solve(int TC) throws Exception{
        int n = ni();
        int B = 400;
        int[] a = new int[1+n];
        for(int i = 1; i<= n; i++)a[i] = ni();
        int[][] ans = new int[1+n][1+B];
        for(int i = n; i>= 1; i--)
            for(int b = 1; b<=B; b++)
                ans[i][b] = (i+a[i]+b>n?1:(1+ans[i+a[i]+b][b]));
        for(int qq = ni(); qq>0; qq--){
            int p = ni(), k = ni();
            if(k<=B)pn(ans[p][k]);
            else{
                int c = 0;
                while(p<=n){
                    p=p+a[p]+k;c++;
                }
                pn(c);
            }
        }
    }
    //SOLUTION ENDS
    long mod = (int)1e9+7, IINF = (long)1e18;
    final int MAX = (int)3e5+1, INF = (int)1e9, root = 3;
    DecimalFormat df = new DecimalFormat("0.000000000000");
    double PI = 3.141592653589793238462643383279502884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        for(int i = 1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{    
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }   
            return str;
        }
    }
}   