#include <bits/stdc++.h>
int n, q, xx, yy, ty, tx, nty, ntx, s, p[110000], plo, v[110000], m[1020],
    e[1020][10], vis[1020][2520], f[1020][2520], clo;
void dfs(int x, int y) {
  int ny = (y + v[x]) % 2520;
  if (ny < 0) ny += 2520;
  int z = e[x][ny % m[x]];
  if (vis[z][ny] && vis[z][ny] < clo) {
    f[x][y] = f[z][ny];
    return;
  }
  if (vis[z][ny] == clo) {
    ty = ny;
    tx = z;
    s = 1;
    p[tx] = ++plo;
    while (1) {
      nty = (ty + v[tx]) % 2520;
      if (nty < 0) nty += 2520;
      ntx = e[tx][nty % m[tx]];
      if ((ntx == z) && (nty == ny)) break;
      tx = ntx;
      if (p[tx] < plo) {
        p[tx] = plo;
        s++;
      }
      ty = nty;
    }
    f[x][y] = s;
    return;
  }
  vis[z][ny] = clo;
  dfs(z, ny);
  f[x][y] = f[z][ny];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &m[i]);
    for (int j = 0; j <= m[i] - 1; j++) scanf("%d", &e[i][j]);
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &xx, &yy);
    yy %= 2520;
    if (yy < 0) yy += 2520;
    if (vis[xx][yy])
      printf("%d\n", f[xx][yy]);
    else {
      vis[xx][yy] = ++clo;
      dfs(xx, yy);
      printf("%d\n", f[xx][yy]);
    }
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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) → … 

### Response

```cpp
#include <bits/stdc++.h>
int n, q, xx, yy, ty, tx, nty, ntx, s, p[110000], plo, v[110000], m[1020],
    e[1020][10], vis[1020][2520], f[1020][2520], clo;
void dfs(int x, int y) {
  int ny = (y + v[x]) % 2520;
  if (ny < 0) ny += 2520;
  int z = e[x][ny % m[x]];
  if (vis[z][ny] && vis[z][ny] < clo) {
    f[x][y] = f[z][ny];
    return;
  }
  if (vis[z][ny] == clo) {
    ty = ny;
    tx = z;
    s = 1;
    p[tx] = ++plo;
    while (1) {
      nty = (ty + v[tx]) % 2520;
      if (nty < 0) nty += 2520;
      ntx = e[tx][nty % m[tx]];
      if ((ntx == z) && (nty == ny)) break;
      tx = ntx;
      if (p[tx] < plo) {
        p[tx] = plo;
        s++;
      }
      ty = nty;
    }
    f[x][y] = s;
    return;
  }
  vis[z][ny] = clo;
  dfs(z, ny);
  f[x][y] = f[z][ny];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &m[i]);
    for (int j = 0; j <= m[i] - 1; j++) scanf("%d", &e[i][j]);
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &xx, &yy);
    yy %= 2520;
    if (yy < 0) yy += 2520;
    if (vis[xx][yy])
      printf("%d\n", f[xx][yy]);
    else {
      vis[xx][yy] = ++clo;
      dfs(xx, yy);
      printf("%d\n", f[xx][yy]);
    }
  }
  return 0;
}

```