#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const long long N = 1005;
long long k[N];
long long m[N];
long long ans[N][2 * 2520];
long long vis[N][2 * 2520];
vector<long long> e[N];
long long normalize(long long x, long long mod = 2520) {
  return ((x % mod) + mod) % mod;
}
long long used[N];
vector<long long> v;
long long dfs(long long x, long long y, long long tim) {
  if (ans[x][y] != -1) {
    return ans[x][y];
  }
  if (vis[x][y] != -1) {
    ans[x][y] = 0;
    for (long long i = vis[x][y]; i < tim; i++) {
      ans[x][y] += !used[v[i]];
      used[v[i]] = 1;
    }
    for (long long i = vis[x][y]; i < tim; i++) {
      used[v[i]] = 0;
    }
    if (ans[x][y] == -1) {
      exit(0);
    }
  }
  if (ans[x][y] == -1) {
    vis[x][y] = tim;
    v.push_back(x);
    long long c = normalize(y + k[x]);
    ans[x][y] = dfs(e[x][c % m[x]], c, tim + 1);
    v.pop_back();
    vis[x][y] = -1;
  }
  return ans[x][y];
}
void solve() {
  memset(ans, -1, sizeof(ans));
  memset(vis, -1, sizeof(vis));
  long long n;
  cin >> n;
  for (long long i = 1; i < n + 1; i++) {
    cin >> k[i];
  }
  for (long long i = 1; i < n + 1; i++) {
    cin >> m[i];
    for (long long j = 0; j < m[i]; j++) {
      long long x;
      cin >> x;
      e[i].push_back(x);
    }
  }
  long long q;
  cin >> q;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    y = normalize(y);
    cout << dfs(x, y, 0) << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const long long N = 1005;
long long k[N];
long long m[N];
long long ans[N][2 * 2520];
long long vis[N][2 * 2520];
vector<long long> e[N];
long long normalize(long long x, long long mod = 2520) {
  return ((x % mod) + mod) % mod;
}
long long used[N];
vector<long long> v;
long long dfs(long long x, long long y, long long tim) {
  if (ans[x][y] != -1) {
    return ans[x][y];
  }
  if (vis[x][y] != -1) {
    ans[x][y] = 0;
    for (long long i = vis[x][y]; i < tim; i++) {
      ans[x][y] += !used[v[i]];
      used[v[i]] = 1;
    }
    for (long long i = vis[x][y]; i < tim; i++) {
      used[v[i]] = 0;
    }
    if (ans[x][y] == -1) {
      exit(0);
    }
  }
  if (ans[x][y] == -1) {
    vis[x][y] = tim;
    v.push_back(x);
    long long c = normalize(y + k[x]);
    ans[x][y] = dfs(e[x][c % m[x]], c, tim + 1);
    v.pop_back();
    vis[x][y] = -1;
  }
  return ans[x][y];
}
void solve() {
  memset(ans, -1, sizeof(ans));
  memset(vis, -1, sizeof(vis));
  long long n;
  cin >> n;
  for (long long i = 1; i < n + 1; i++) {
    cin >> k[i];
  }
  for (long long i = 1; i < n + 1; i++) {
    cin >> m[i];
    for (long long j = 0; j < m[i]; j++) {
      long long x;
      cin >> x;
      e[i].push_back(x);
    }
  }
  long long q;
  cin >> q;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    y = normalize(y);
    cout << dfs(x, y, 0) << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```