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.awt.*;
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintStream;
import java.util.*;

public class CF_1200F {
    public static void main(String args[]) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        PrintStream out = new PrintStream(new BufferedOutputStream(System.out));

        int lcm = 8 * 9 * 5 * 7;

        int n = in.nextInt();
        int[] k = new int[n];
        for (int i = 0; i < n; i++) k[i] = Math.floorMod(in.nextInt(), lcm);

        int[][] adj = new int[n][];
        for (int i = 0; i < n; i++) {
            int m = in.nextInt();
            adj[i] = new int[m];

            for (int j = 0; j < m; j++) adj[i][j] = in.nextInt() - 1;
        }

        int q = in.nextInt();

        int[][] dp = new int[n][lcm];
        int[][] iLp = new int[n][lcm];
        for (int[] a : dp) Arrays.fill(a, -1);
        for (int[] a : iLp) Arrays.fill(a, -1);
        for (int i = 0; i < q; i++) {
            int x = in.nextInt() - 1, y = Math.floorMod(in.nextInt() + k[x], lcm);

            ArrayList<Point> hist = new ArrayList<>();
            while (dp[x][y] == -1 && iLp[x][y] == -1) {
                iLp[x][y] = hist.size();
                hist.add(new Point(x, y));

                x = adj[x][y % adj[x].length];
                y = Math.floorMod(y + k[x], lcm);
            }


            int ret;
            if (dp[x][y] == -1) {
                HashSet<Integer> vis = new HashSet<>();
                for (int j = iLp[x][y]; j < hist.size(); j++) {
                    vis.add(hist.get(j).x);
                }
                ret = vis.size();
            } else {
                ret = dp[x][y];
            }

            for (Point p : hist) {
                dp[p.x][p.y] = ret;
            }

            out.println(dp[x][y]);
        }


        out.close();
        in.close();
    }
}
