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;
template <typename T, typename L>
bool smax(T &x, L y) {
  return x < y ? (x = y, true) : false;
}
template <typename T, typename L>
bool smin(T &x, L y) {
  return y < x ? (x = y, true) : false;
}
int root(int u, int c);
void merge(int u, int v, int c);
const int MAXN = 1e5 + 15;
unordered_map<int, int> par[MAXN], dp[MAXN];
int n, m, q;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cerr << "HELLO WORLD :)\n";
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    merge(a, b, c);
  }
  cin >> q;
  while (q--) {
    int u, v;
    cin >> u >> v;
    if (par[u].size() > par[v].size()) swap(u, v);
    if (!dp[u].count(v)) {
      int &res = dp[u][v];
      for (auto c : par[u])
        if (par[v].count(c.first) && root(v, c.first) == root(u, c.first))
          res++;
    }
    cout << dp[u][v] << '\n';
  }
  return false;
}
int root(int u, int c) {
  return (par[u][c] < 0 ? u : par[u][c] = root(par[u][c], c));
}
void merge(int u, int v, int c) {
  if (!par[u].count(c)) par[u][c] = -1;
  if (!par[v].count(c)) par[v][c] = -1;
  u = root(u, c);
  v = root(v, c);
  if (v == u) return;
  if (par[v][c] > par[u][c]) swap(u, v);
  par[v][c] += par[u][c];
  par[u][c] = v;
}
