Mr. Kitayuta has just bought an undirected graph with n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color ci, connecting vertex ai and bi.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers - ui and vi.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex ui and vertex vi directly or indirectly.

Input

The first line of the input contains space-separated two integers - n and m(2 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers - ai, bi(1 ≤ ai < bi ≤ n) and ci(1 ≤ ci ≤ m). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if i ≠ j, (ai, bi, ci) ≠ (aj, bj, cj).

The next line contains a integer- q(1 ≤ q ≤ 105), denoting the number of the queries.

Then follows q lines, containing space-separated two integers - ui and vi(1 ≤ ui, vi ≤ n). It is guaranteed that ui ≠ vi.

Output

For each query, print the answer in a separate line.

Examples

Input

4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4


Output

2
1
0


Input

5 7
1 5 1
2 5 1
3 5 1
4 5 1
1 2 2
2 3 2
3 4 2
5
1 5
5 1
2 5
1 5
1 4


Output

1
1
1
1
2

Note

Let's consider the first sample. 

<image> The figure above shows the first sample. 

  * Vertex 1 and vertex 2 are connected by color 1 and 2. 
  * Vertex 3 and vertex 4 are connected by color 3. 
  * Vertex 1 and vertex 4 are not connected by any single color. 
#include <bits/stdc++.h>
using namespace std;
const long double EPS = 10e-9;
const int INF = 1e9 + 5;
const long long INFLL = (long long)INF * (long long)INF;
const long long MOD = 1e9 + 7;
const long double PI = 3.14159265358979323846;
struct HASH {
  size_t operator()(const pair<int, int>& x) const {
    return hash<long long>()(((long long)x.first) ^
                             (((long long)x.second) << 32));
  }
};
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = 0; i < (int)v.size(); ++i) {
    os << v[i];
    if (i != (int)v.size() - 1) os << ", ";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "{";
  for (const auto& it : v) {
    os << it;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "}\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const unordered_set<T>& v) {
  os << "{";
  for (const auto& it : v) {
    os << it;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "}\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  os << "{";
  for (const auto& it : v) {
    os << it.first << " : " << it.second;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "}\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const unordered_map<T, S>& v) {
  os << "{";
  for (const auto& it : v) {
    os << it.first << " : " << it.second;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "}\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << "(";
  os << v.first << ", " << v.second << ")";
  os << "\n";
  return os;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
bool dfs(vector<vector<int>>& g, int start, int end, vector<bool>& visited) {
  if (start == end) {
    return true;
  }
  visited[start] = true;
  bool ret = false;
  for (int n : g[start]) {
    if (!visited[n]) {
      ret = ret || dfs(g, n, end, visited);
    }
  }
  return ret;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<vector<int>>> g(m + 1,
                                vector<vector<int>>(n + 1, vector<int>()));
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    g[c][a].push_back(b);
    g[c][b].push_back(a);
  }
  vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
  for (int u = 1; u <= n; u++) {
    for (int v = 1; v <= n; v++) {
      if (dp[u][v]) {
        continue;
      }
      for (int c = 1; c <= m; c++) {
        vector<bool> visited(n + 1, false);
        bool f = dfs(g[c], u, v, visited);
        if (f) {
          dp[u][v]++;
        }
      }
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    cout << dp[u][v] << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
