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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author beginner1010
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskF solver = new TaskF();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskF {
        final int LCM = 9 * 8 * 7 * 5;

        int getMod(int val) {
            val %= LCM;
            if (val < 0) val += LCM;
            return val;
        }

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int[] K = new int[n];
            for (int i = 0; i < n; i++) {
                K[i] = getMod(in.nextInt());
            }

            int[] next = new int[n * LCM];
            for (int i = 0; i < n; i++) {
                int m = in.nextInt();
                int[] edges = new int[m];
                for (int j = 0; j < m; j++) {
                    edges[j] = in.nextInt() - 1;
                }
                for (int y = 0; y < LCM; y++) {
                    int c = getMod(y + K[i]);
                    int neighborID = c % m;
                    next[y + i * LCM] = edges[neighborID] * LCM + c;
                }
            }

            int nodes = n * LCM;
            int[] vis = new int[n * LCM];
            int[] ans = new int[n * LCM];
            int[] posInPath = new int[n * LCM];
            int pathID = 0;
            for (int i = 0; i < nodes; i++) {
                if (vis[i] == 0) {
                    pathID++;
                    int curNode = i;
                    ArrayList<Integer> path = new ArrayList<>();
                    while (vis[curNode] == 0) {
                        vis[curNode] = pathID;
                        posInPath[curNode] = path.size();
                        path.add(curNode);
                        curNode = next[curNode];
                    }
                    if (vis[curNode] == pathID) {
                        HashSet<Integer> set = new HashSet<>();
                        for (int j = posInPath[curNode]; j < path.size(); j++) {
                            set.add(path.get(j) / LCM);
                        }
                        ans[curNode] = set.size();
                    }
                    for (int node : path) {
                        ans[node] = ans[curNode];
                    }
                }
            }
            int q = in.nextInt();
            for (int i = 0; i < q; i++) {
                int x = in.nextInt() - 1;
                int y = getMod(in.nextInt());
                out.println(ans[x * LCM + y]);
            }
        }

    }

    static class InputReader {
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private InputStream stream;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        private boolean isWhitespace(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        private int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = read();
            while (isWhitespace(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isWhitespace(c));
            return res * sgn;
        }

    }
}

