#include <bits/stdc++.h>
using namespace std;
int xm[4] = {-1, 1, 0, 0};
int ym[4] = {0, 0, -1, 1};
const int MOD = 1e9 + 7;
const int K = 2520;
const int MAXN = 1000 + 5;
const long long POW = 9973;
int dp[1005][K + 5];
vector<int> adj[MAXN];
int ar[MAXN];
int vis[1005][K];
void form(int &b) { b = ((b % K) + K) % K; }
void dfs(int x, int k) {
  vector<pair<int, int>> v;
  int ans = 0;
  int flg = 0;
  form(k);
  v.push_back(make_pair(x, k));
  pair<int, int> end;
  int nx, nk, m;
  while (true) {
    vis[x][k] = 1;
    m = adj[x].size();
    nx = adj[x][k % m];
    nk = k + ar[nx];
    form(nk);
    v.push_back(make_pair(nx, nk));
    if (dp[nx][nk]) {
      ans = dp[nx][nk];
      break;
    }
    if (vis[nx][nk]) {
      break;
    }
    x = nx;
    k = nk;
  }
  set<int> distinct;
  if (!ans) {
    for (pair<int, int> p : v) {
      if (p == make_pair(nx, nk)) flg = 1;
      if (flg) distinct.insert(p.first);
    }
    ans = distinct.size();
  }
  for (pair<int, int> p : v) {
    dp[p.first][p.second] = ans;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> ar[i];
  for (int i = 1; i <= n; i++) {
    int m;
    cin >> m;
    for (int j = 0; j < m; j++) {
      int x;
      cin >> x;
      adj[i].push_back(x);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < K; j++) {
      if (dp[i][j]) continue;
      dfs(i, j);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b;
    cin >> a >> b;
    b += ar[a];
    form(b);
    cout << dp[a][b] << "\n";
  }
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
int xm[4] = {-1, 1, 0, 0};
int ym[4] = {0, 0, -1, 1};
const int MOD = 1e9 + 7;
const int K = 2520;
const int MAXN = 1000 + 5;
const long long POW = 9973;
int dp[1005][K + 5];
vector<int> adj[MAXN];
int ar[MAXN];
int vis[1005][K];
void form(int &b) { b = ((b % K) + K) % K; }
void dfs(int x, int k) {
  vector<pair<int, int>> v;
  int ans = 0;
  int flg = 0;
  form(k);
  v.push_back(make_pair(x, k));
  pair<int, int> end;
  int nx, nk, m;
  while (true) {
    vis[x][k] = 1;
    m = adj[x].size();
    nx = adj[x][k % m];
    nk = k + ar[nx];
    form(nk);
    v.push_back(make_pair(nx, nk));
    if (dp[nx][nk]) {
      ans = dp[nx][nk];
      break;
    }
    if (vis[nx][nk]) {
      break;
    }
    x = nx;
    k = nk;
  }
  set<int> distinct;
  if (!ans) {
    for (pair<int, int> p : v) {
      if (p == make_pair(nx, nk)) flg = 1;
      if (flg) distinct.insert(p.first);
    }
    ans = distinct.size();
  }
  for (pair<int, int> p : v) {
    dp[p.first][p.second] = ans;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> ar[i];
  for (int i = 1; i <= n; i++) {
    int m;
    cin >> m;
    for (int j = 0; j < m; j++) {
      int x;
      cin >> x;
      adj[i].push_back(x);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < K; j++) {
      if (dp[i][j]) continue;
      dfs(i, j);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b;
    cin >> a >> b;
    b += ar[a];
    form(b);
    cout << dp[a][b] << "\n";
  }
}

```