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 int N = 1e5 + 2;
struct Pair {
  int a, b;
};
struct dsu {
  vector<int> par;
  dsu(int n = 0) {
    par.resize(n);
    for (int i = 0; i < n; ++i) par[i] = i;
  }
  int get(int x) { return (x == par[x] ? x : par[x] = get(par[x])); }
  void unite(int x, int y) {
    x = get(x), y = get(y);
    par[y] = x;
  }
};
int n, m, q, Bucket_sz;
int q_ans[1 + N];
vector<Pair> E[1 + N];
Pair query[1 + N];
map<pair<int, int>, vector<int>> pos_q;
void read() {
  scanf("%d%d", &n, &m);
  Bucket_sz = sqrt(m);
  for (int i = 1; i <= m; ++i) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    E[c].push_back(Pair{a, b});
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &query[i].a, &query[i].b);
    if (query[i].a > query[i].b) swap(query[i].a, query[i].b);
    pos_q[make_pair(query[i].a, query[i].b)].push_back(i);
  }
}
void solve_big() {
  dsu forest(n + 1);
  for (int i = 1; i <= m; ++i) {
    if ((int)E[i].size() >= Bucket_sz) {
      forest = dsu(n + 1);
      for (auto edges : E[i]) forest.unite(edges.a, edges.b);
      for (int j = 1; j <= q; ++j)
        if (forest.get(query[j].a) == forest.get(query[j].b)) q_ans[j]++;
    }
  }
}
void solve_small() {
  dsu forest(n + 1);
  for (int c = 1; c <= m; ++c) {
    if ((int)E[c].size() < Bucket_sz && !E[c].empty()) {
      for (auto edges : E[c]) forest.unite(edges.a, edges.b);
      vector<int> nodes;
      for (auto edges : E[c])
        nodes.push_back(edges.a), nodes.push_back(edges.b);
      sort(nodes.begin(), nodes.end());
      nodes.resize(unique(nodes.begin(), nodes.end()) - nodes.begin());
      for (int i = 0; i < (int)nodes.size(); ++i)
        for (int j = i + 1; j < (int)nodes.size(); ++j)
          if (forest.get(nodes[i]) == forest.get(nodes[j]))
            if (pos_q.find(make_pair(nodes[i], nodes[j])) != end(pos_q))
              for (auto pozitii : pos_q[make_pair(nodes[i], nodes[j])])
                q_ans[pozitii]++;
      for (auto nod : nodes) forest.par[nod] = nod;
    }
  }
}
void solve() {
  solve_big();
  solve_small();
  for (int i = 1; i <= q; ++i) printf("%d\n", q_ans[i]);
}
int main() {
  read();
  solve();
}
