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.util.*;
import java.io.*;

public class GraphTraveler {
    static int[][] edges;
    static int[] nums;
    static int[][] dp;
    static int mod = 2520;
    static int[][] depth;
    static int[][] color;
    static ArrayList<Integer> prev;
    static ArrayList<Integer> curc;
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int n = scanner.nextInt();
        edges = new int[n][];
        nums = new int[n];
        dp = new int[n][mod];
        color = new int[n][mod];
        depth = new int[n][mod];
        for(int i = 0; i < n; i++) nums[i] = scanner.nextInt();
        for(int i = 0; i < n; i++) {
            int m = scanner.nextInt();
            int[] edge = new int[m];
            Arrays.fill(dp[i], -1);
            for(int j = 0; j < m; j++) edge[j] = scanner.nextInt()-1;
            edges[i] = edge;
        }
        int q = scanner.nextInt();
        while(q-->0) {
            int v = scanner.nextInt()-1;
            int c = ((scanner.nextInt()%mod) + mod) % mod;
            if (dp[v][c] != -1) out.println(dp[v][c]);
            else {
                curc = new ArrayList<>();
                prev = new ArrayList<>();
                int d = 0;
                while(true) {
                    curc.add(c);
                    prev.add(v);
                    depth[v][c] = d;
                    color[v][c] = q+1;
                    int nextc = (((c + nums[v]) % mod) + mod) % mod;
                    int upd = (((c + nums[v]) % edges[v].length) + edges[v].length) % edges[v].length;
                    int next = edges[v][upd];
                    if (dp[next][nextc] == -1) {
                        if (color[next][nextc] == q+1) {
                            int dd = d;
                            HashSet<Integer> seen = new HashSet<>();
                            seen.add(next);
                            while(dd > 0 && (prev.get(dd) != next || curc.get(dd)!= nextc)) {
                                seen.add(prev.get(dd));
                                dd--;
                            }
                            dp[v][c] = seen.size();
                            while(d>0) {
                                dp[prev.get(d-1)][curc.get(d-1)] = dp[v][c];
                                c = curc.get(d-1);
                                v = prev.get(d-1);;
                                d--;
                            }
                            break;
                        }
                        else {
                            v = next;
                            c = nextc;
                        }
                    }
                    else {
                        dp[v][c]=dp[next][nextc];
                        while(d>0) {
                            dp[prev.get(d-1)][curc.get(d-1)] = dp[v][c];
                            c = curc.get(d-1);
                            v = prev.get(d-1);;
                            d--;
                        }
                        break;
                    }
                    d++;
                }
                //out.println(Arrays.toString(curc.toArray()));
                //out.println(Arrays.toString(prev.toArray()));
                out.println(dp[prev.get(0)][curc.get(0)]);
            }
        }
        out.flush();
    }
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        
        public FastScanner() {
            this(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
        
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}
