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 MAXN = 100001;
int n, m, q;
vector<pair<int, int> > color[MAXN];
map<pair<int, int>, int> Q;
map<int, int> p[MAXN];
vector<pair<int, int> > comp[MAXN];
set<int> colors[MAXN];
int root(int a, int c) {
  if (p[c].find(a) == p[c].end()) return p[c][a] = a;
  int &x = p[c][a];
  return x == a ? x : x = root(x, c);
}
void merge(int a, int b, int c) {
  int aa = root(a, c);
  int bb = root(b, c);
  if (aa == bb) return;
  p[c][aa] = bb;
  colors[a].insert(c);
  colors[b].insert(c);
}
int solve(int a, int b) {
  int res = -1;
  if (comp[a].size() > comp[b].size()) swap(a, b);
  for (int i = 0; i < (int)comp[a].size(); ++i)
    res +=
        comp[a][i] == *lower_bound(comp[b].begin(), comp[b].end(), comp[a][i]);
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, a, b, c; i < m; ++i) {
    scanf("%d%d%d", &a, &b, &c);
    merge(a, b, c);
  }
  for (int i = 1; i <= n; ++i) {
    for (set<int>::iterator it = colors[i].begin(); it != colors[i].end(); ++it)
      comp[i].push_back(make_pair(*it, root(i, *it)));
    sort(comp[i].begin(), comp[i].end());
    comp[i].push_back(make_pair(1e9, 1e9));
  }
  scanf("%d", &q);
  for (int i = 0, a, b; i < q; ++i) {
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    if (Q.find(make_pair(a, b)) == Q.end()) Q[make_pair(a, b)] = solve(a, b);
    printf("%d\n", Q[make_pair(a, b)]);
  }
  return 0;
}
