Gildong is experimenting with an interesting machine Graph Traveler. In Graph Traveler, there is a directed graph consisting of n vertices numbered from 1 to n. The i-th vertex has m_i outgoing edges that are labeled as e_i[0], e_i[1], …, e_i[m_i-1], each representing the destination vertex of the edge. The graph can have multiple edges and self-loops. The i-th vertex also has an integer k_i written on itself.

A travel on this graph works as follows. 

  1. Gildong chooses a vertex to start from, and an integer to start with. Set the variable c to this integer. 
  2. After arriving at the vertex i, or when Gildong begins the travel at some vertex i, add k_i to c. 
  3. The next vertex is e_i[x] where x is an integer 0 ≤ x ≤ m_i-1 satisfying x ≡ c \pmod {m_i}. Go to the next vertex and go back to step 2. 



It's obvious that a travel never ends, since the 2nd and the 3rd step will be repeated endlessly.

For example, assume that Gildong starts at vertex 1 with c = 5, and m_1 = 2, e_1[0] = 1, e_1[1] = 2, k_1 = -3. Right after he starts at vertex 1, c becomes 2. Since the only integer x (0 ≤ x ≤ 1) where x ≡ c \pmod {m_i} is 0, Gildong goes to vertex e_1[0] = 1. After arriving at vertex 1 again, c becomes -1. The only integer x satisfying the conditions is 1, so he goes to vertex e_1[1] = 2, and so on.

Since Gildong is quite inquisitive, he's going to ask you q queries. He wants to know how many distinct vertices will be visited infinitely many times, if he starts the travel from a certain vertex with a certain value of c. Note that you should not count the vertices that will be visited only finite times.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 1000), the number of vertices in the graph.

The second line contains n integers. The i-th integer is k_i (-10^9 ≤ k_i ≤ 10^9), the integer written on the i-th vertex.

Next 2 ⋅ n lines describe the edges of each vertex. The (2 ⋅ i + 1)-st line contains an integer m_i (1 ≤ m_i ≤ 10), the number of outgoing edges of the i-th vertex. The (2 ⋅ i + 2)-nd line contains m_i integers e_i[0], e_i[1], …, e_i[m_i-1], each having an integer value between 1 and n, inclusive.

Next line contains an integer q (1 ≤ q ≤ 10^5), the number of queries Gildong wants to ask.

Next q lines contains two integers x and y (1 ≤ x ≤ n, -10^9 ≤ y ≤ 10^9) each, which mean that the start vertex is x and the starting value of c is y.

Output

For each query, print the number of distinct vertices that will be visited infinitely many times, if Gildong starts at vertex x with starting integer y.

Examples

Input


4
0 0 0 0
2
2 3
1
2
3
2 4 1
4
3 1 2 1
6
1 0
2 0
3 -1
4 -2
1 1
1 5


Output


1
1
2
1
3
2


Input


4
4 -5 -3 -1
2
2 3
1
2
3
2 4 1
4
3 1 2 1
6
1 0
2 0
3 -1
4 -2
1 1
1 5


Output


1
1
1
3
1
1

Note

The first example can be shown like the following image:

<image>

Three integers are marked on i-th vertex: i, k_i, and m_i respectively. The outgoing edges are labeled with an integer representing the edge number of i-th vertex.

The travel for each query works as follows. It is described as a sequence of phrases, each in the format "vertex (c after k_i added)".

  * 1(0) → 2(0) → 2(0) → … 
  * 2(0) → 2(0) → … 
  * 3(-1) → 1(-1) → 3(-1) → … 
  * 4(-2) → 2(-2) → 2(-2) → … 
  * 1(1) → 3(1) → 4(1) → 1(1) → … 
  * 1(5) → 3(5) → 1(5) → … 



The second example is same as the first example, except that the vertices have non-zero values. Therefore the answers to the queries also differ from the first example.

<image>

The queries for the second example works as follows:

  * 1(4) → 2(-1) → 2(-6) → … 
  * 2(-5) → 2(-10) → … 
  * 3(-4) → 1(0) → 2(-5) → 2(-10) → … 
  * 4(-3) → 1(1) → 3(-2) → 4(-3) → … 
  * 1(5) → 3(2) → 1(6) → 2(1) → 2(-4) → … 
  * 1(9) → 3(6) → 2(1) → 2(-4) → … 

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.StringTokenizer;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Andy Phan
 */

public class f implements Runnable {
    static int MOD = 2520;
    static long BIG = 1987654321l;
    static int[] dp;
    static int[] next;
    
    public static void main(String[] args) {
        new Thread(null, new f(), "lol", 1l<<28).start();
    }

    @Override
    public void run() {
        FS in = new FS(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
        int n = in.nextInt();
        
        long[] val = new long[n];
        
        for(int i = 0; i < n; i++) val[i] = in.nextInt();
        int[][] edges = new int[n][];
        
        for(int i = 0; i < n; i++) {
            int m = in.nextInt();
            edges[i] = new int[m];
            for(int j = 0; j < m; j++) edges[i][j] = in.nextInt()-1;
        }
        
        next = new int[MOD*n];
        for(int i = 0; i < MOD; i++) {
            for(int j = 0; j < n; j++) {
                int id = j*MOD+i;
                int nextVal = (int)((val[j]+MOD*BIG+i)%MOD);
                int nextNode = edges[j][nextVal%edges[j].length];
                next[id] = nextNode*MOD+nextVal;
            }
        }
        dp = new int[MOD*n];
        Arrays.fill(dp, -1);
        
        int q = in.nextInt();
        while(q-->0) {
            int x = in.nextInt()-1;
            int y = in.nextInt();
            y = (int)((y+MOD*BIG)%MOD);
            
            ArrayDeque<Integer> qe = new ArrayDeque<>();
            qe.add(x*MOD+y);
            int res = -1;
            while(!qe.isEmpty()) {
                int node = qe.pollLast();
                if(res != -1) {
                    dp[node] = res;
                    continue;
                }
                if(dp[node] >= 0) {
                    res = dp[node];
                    continue;
                }
                if(next[node] == node) {
                    res = dp[node] = 1;
                    continue;
                }
                if(dp[node] == -2) {
                    HashSet<Integer> set = new HashSet<>();
                    int curr = node;
                    do {
                        set.add(curr/MOD);
                        curr = next[curr];
                    } while(curr != node);
                    do {
                        dp[curr] = set.size();
                    } while(curr != node);
                    res = dp[curr];
                    continue;
                }
                dp[node] = -2;
                qe.addLast(node);
                qe.addLast(next[node]);
            }
            System.out.println(res);
        }
        
        out.close();
    }
    static class FS {

        BufferedReader in;
        StringTokenizer token;
        
        public FS(InputStream str) {
            in = new BufferedReader(new InputStreamReader(str));
        }
        
        public String next() {
            if (token == null || !token.hasMoreElements()) {
                try {
                    token = new StringTokenizer(in.readLine());
                } catch (IOException ex) {
                }
                return next();
            }
            return token.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}
