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 B = 320;
const int N = 100000;
const int HASHSIZE = 1000037;
int ans[N + 5], n, m, q;
vector<pair<int, int> > edges[N + 5];
vector<pair<pair<int, int>, int> > hsh[HASHSIZE + 5];
int func(const pair<int, int> &x) {
  return (1ll * x.first * 1313131 + x.second) % HASHSIZE;
}
void hinsert(const pair<int, int> &x, const int &i) {
  hsh[func(x)].push_back(make_pair(x, i));
}
void hsearch(const pair<int, int> &x) {
  int k = func(x);
  for (auto p : hsh[k])
    if (x == p.first) ans[p.second]++;
}
int fa[N + 5];
bool vis[N + 5];
int find(int x) { return fa[x] = (fa[x] == x ? x : find(fa[x])); }
bool unite(int x, int y) {
  int fx = find(x), fy = find(y);
  if (fx != fy) {
    fa[fx] = fy;
    return true;
  } else
    return false;
}
int x[N + 5], y[N + 5];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, u, v, c; i <= m; i++)
    scanf("%d%d%d", &u, &v, &c), edges[c].push_back(make_pair(u, v));
  scanf("%d", &q);
  for (int i = 1; i <= q; i++)
    scanf("%d%d", &x[i], &y[i]),
        hinsert(make_pair(min(x[i], y[i]), max(x[i], y[i])), i);
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= m; i++) {
    if ((int)edges[i].size() <= B) {
      vector<int> vec;
      for (auto e : edges[i]) {
        if (!vis[e.first]) vec.push_back(e.first), vis[e.first] = true;
        if (!vis[e.second]) vec.push_back(e.second), vis[e.second] = true;
        unite(e.first, e.second);
      }
      for (unsigned u = 0; u < vec.size(); u++)
        for (unsigned v = u + 1; v < vec.size(); v++)
          if (find(vec[u]) == find(vec[v]))
            hsearch(make_pair(min(vec[u], vec[v]), max(vec[u], vec[v])));
      for (auto k : vec) vis[k] = false, fa[k] = k;
    } else {
      for (auto e : edges[i]) unite(e.first, e.second);
      for (int j = 1; j <= q; j++)
        if (find(x[j]) == find(y[j])) ans[j]++;
      for (int j = 1; j <= n; j++) fa[j] = j;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
