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.*;
import java.util.*;

public class Sol implements Runnable{
    public static void main(String[] args) {
        new Thread(null, new Sol(), "process", 1<<26).start();
    }
	public void run() {
		FastReader scan = new FastReader();
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		//PrintWriter out = new PrintWriter("file.out");
		Task solver = new Task();
		//int t = scan.nextInt();
		int t = 1;
		for(int i = 1; i <= t; i++) solver.solve(i, scan, out);
		out.close();
	}

	static class Task {
		static final int inf = Integer.MAX_VALUE;
		final public static int MOD = 2520;
		public static int k[];
		public static int n;
		public static List<Integer>adj[];
		public static int m[];
		public static int dp[][] = new int[1000][2520];
		public static int vis[][] = new int[1000][2520];
		public static int use[] = new int[1000];
		public static int idx, id;
		public void solve(int testNumber, FastReader sc, PrintWriter pw) {
			n = sc.nextInt();
			k = new int[n];
			adj = new ArrayList[n];
			m = new int[n];
			for(int i=0; i<n; i++) {
				k[i] = sc.nextInt();
				adj[i] = new ArrayList<>();
			}
			for(int i=0; i<n; i++) {
				m[i] = sc.nextInt();
				for(int j=0; j<m[i]; j++) {
					int a = sc.nextInt()-1;
					adj[i].add(a);
				}
			}
			for(int i=0; i<1000; i++) {
				Arrays.fill(dp[i], -1);
			}
			int q = sc.nextInt();
			int ans[] = new int[q];
			Arrays.fill(ans, 0);
			for(int i=0; i<q; i++) {
				idx++;
				int start = sc.nextInt()-1;
				int value = sc.nextInt();
				ArrayDeque<pair> ad = new ArrayDeque<>();
				ad.add(new pair(start, value));
				while(!ad.isEmpty()) {
					pair p = ad.poll();
					int curr = p.a;
					int val = p.b;
					val = (val+k[curr])%MOD;
					if(val<0) val+=MOD;
					if(dp[curr][val]!=-1&&ans[dp[curr][val]]!=0) {
						ans[i] = ans[dp[curr][val]];
						break;
					}
					dp[curr][val] = i;
					if(vis[curr][val]==idx) {
						ans[i] = DFS2(curr, (int)val, adj[curr].get((int)val%m[curr]), (int)val, 0);
						break;
					}
					vis[curr][val] = idx;
					ad.add(new pair(adj[curr].get(val%m[curr]), val));
				}
				pw.println(ans[i]);
			}
			pw.close();
		}
		/*public boolean visited(int curr, int val) {
			val = val + k[curr];
			val%=MOD;
			if(val<0) val+=MOD;
			if(vis[curr][val]==idx) return true;
			return false;
		}
		public static int DFS(int curr, long val) {
			val = val+k[curr];
			val%=MOD;
			if(val<0) val+=MOD;
			if(dp[curr][(int)val]!=0) {
				return dp[curr][(int)val];
			}
			if(vis[curr][(int)val]==idx) {
				int good = DFS2(curr, (int)val, adj[curr].get((int)val%m[curr]), (int)val, 0);
				return dp[curr][(int)val] = good;
			}
			vis[curr][(int)val] = idx;
			return DFS(adj[curr].get((int)val%m[curr]), (int)val);
		}
		public static void DFS1(int curr, long val, int ans) {
			val = val+k[curr];
			val%=MOD;
			if(val<0) val+=MOD;
			dp[curr][(int)val] = ans;
			if(vis[curr][(int)val]==id) {
				return;
			}
			vis[curr][(int)val] = id;
			DFS1(adj[curr].get((int)val%m[curr]), val, ans);
		}*/
		public static int DFS2(int a, int b, int current, int value, int cnt) {
			ArrayDeque<pair> ad = new ArrayDeque<>();
			ad.add(new pair(current, value));
			while(true) {
				pair p = ad.poll();
				int curr = p.a;
				int val = p.b;
				val = (val+k[curr])%MOD;
				if(val<0) val+=MOD;
				if(use[curr]!=idx) cnt++;
				use[curr] = idx;
				if(curr==a&&val==b) return cnt;
				ad.add(new pair(adj[curr].get(val%m[curr]), val));
			}
		}
	}
	static long binpow(long a, long b, long m) {
		a %= m;
		long res = 1;
		while (b > 0) {
			if ((b & 1) == 1)
				res = res * a % m;
			a = a * a % m;
			b >>= 1;
		}
		return res;
	}
	static class pair {
		int a,b;
		public pair(int a, int b) {
			this.a = a;
			this.b = b;
		}
	}
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

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

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

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

}