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) → … 
#include <bits/stdc++.h>
using namespace std;
const long long lcm = 2520;
long long a[1001];
long long m[1001];
long long out[1001][11];
long long tot;
long long vis[1001][2520];
long long ans[1001][2520];
unordered_map<int, bool> viss;
long long xx, yyy, fst;
long long ct(long long px, long long py) {
  while (true) {
    if (!viss[int(px)]) viss[int(px)] = true, tot++;
    if (fst && px == xx && py == yyy) return tot;
    py += a[px], py += 2520000000, py %= lcm;
    fst = 1;
    px = out[px][py % m[px]];
  }
  return tot;
}
long long dfs(long long x, long long y) {
  long long yy = y;
  if (vis[x][y]) {
    if (ans[x][y] == 0) {
      xx = x, yyy = y, fst = 0;
      return ct(x, y);
    } else
      return ans[x][y];
  }
  y += a[x], y += 2520000000, y %= lcm;
  vis[x][yy] = 1;
  ans[x][yy] = dfs(out[x][y % m[x]], y);
  return ans[x][yy];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  long long i, j;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    cin >> m[i];
    for (j = 0; j < m[i]; j++) cin >> out[i][j];
  }
  for (i = 1; i <= n; i++)
    for (j = 0; j < lcm; j++)
      if (!vis[i][j]) tot = 0, viss.clear(), dfs(i, j);
  cin >> q;
  for (i = 1; i <= q; i++) {
    long long x, y;
    cin >> x >> y;
    cout << ans[x][(y + 2520000000) % lcm] << endl;
  }
  return 0;
}
