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 INF = 1e9 + 7;
struct UnionFind {
  vector<int> parent;
  UnionFind(int n) : parent(n, -1) {}
  int root(int x) { return parent[x] < 0 ? x : parent[x] = root(parent[x]); }
  bool merge(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return false;
    if (parent[y] < parent[x]) swap(x, y);
    if (parent[x] == parent[y]) --parent[x];
    parent[y] = x;
    return true;
  }
};
int N, M, Q;
int u[100000], v[100000];
vector<pair<int, int>> es[100000];
int res[100000];
map<pair<int, int>, vector<int>> ma;
pair<int, int> f(int u, int v) { return pair<int, int>(min(u, v), max(u, v)); }
int main(void) {
  cin >> N >> M;
  for (int i = 0; i < (int)(M); ++i) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    a--, b--, c--;
    es[c].push_back(pair<int, int>(a, b));
  }
  cin >> Q;
  for (int i = 0; i < (int)(Q); ++i) {
    scanf("%d %d", u + i, v + i);
    u[i]--, v[i]--;
    ma[f(u[i], v[i])].push_back(i);
  }
  for (int c = 0; c < (int)(M); ++c) {
    if ((int)es[c].size() <= 300) {
      vector<int> rev;
      for (pair<int, int> p : es[c])
        rev.push_back(p.first), rev.push_back(p.second);
      sort(rev.begin(), rev.end());
      rev.erase(unique(rev.begin(), rev.end()), rev.end());
      int NN = rev.size();
      map<int, int> cmp;
      for (int i = 0; i < (int)(NN); ++i) cmp[rev[i]] = i;
      UnionFind uf(NN);
      for (pair<int, int> p : es[c]) {
        int u = cmp[p.first], v = cmp[p.second];
        uf.merge(u, v);
      }
      for (int j = 0; j < (int)(NN); ++j)
        for (int i = 0; i < (int)(j); ++i)
          if (uf.root(i) == uf.root(j))
            for (int id : ma[f(rev[i], rev[j])]) res[id]++;
    } else {
      UnionFind uf(N);
      for (pair<int, int> p : es[c]) uf.merge(p.first, p.second);
      for (int i = 0; i < (int)(Q); ++i)
        if (uf.root(u[i]) == uf.root(v[i])) res[i]++;
    }
  }
  for (int i = 0; i < (int)(Q); ++i) cout << res[i] << endl;
  return 0;
}
