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;
int n, m, q, ccnt[100005], fth[100005 << 1];
vector<int> colo[100005];
pair<pair<int, int>, int> info[100005];
int rarg(vector<int> &a) {
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
  return a.size();
}
int find(int a, int c) {
  return ccnt[a - 1] + lower_bound(colo[a].begin(), colo[a].end(), c) -
         colo[a].begin();
}
void init() {
  for (int i = 0; i < ccnt[n]; ++i) fth[i] = i;
}
int getf(int a) { return fth[a] == a ? a : (fth[a] = getf(fth[a])); }
void merge(int A, int B) { fth[getf(A)] = getf(B); }
map<pair<int, int>, int> memo;
int ccount(int a, int b) {
  int ans = 0, l = colo[a].size();
  for (int i = 0; i < l; ++i) {
    int c = colo[a][i];
    if (binary_search(colo[b].begin(), colo[b].end(), c) &&
        getf(find(a, c)) == getf(find(b, c)))
      ans++;
  }
  memo[make_pair(a, b)] = ans;
  memo[make_pair(b, a)] = ans;
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, a, b, c; i < m; ++i) {
    scanf("%d%d%d", &a, &b, &c);
    colo[a].push_back(c), colo[b].push_back(c);
    info[i] = make_pair(make_pair(a, b), c);
  }
  for (int i = 1; i <= n; ++i) ccnt[i] = rarg(colo[i]) + ccnt[i - 1];
  init();
  for (int i = 0, a, b, c, A, B; i < m; ++i) {
    a = info[i].first.first;
    b = info[i].first.second;
    c = info[i].second;
    A = find(a, c), B = find(b, c);
    merge(A, B);
  }
  scanf("%d", &q);
  for (int i = 0, a, b; i < q; ++i) {
    scanf("%d%d", &a, &b);
    pair<int, int> p = make_pair(a, b);
    if (memo.count(p))
      printf("%d\n", memo[p]);
    else {
      if (colo[a].size() > colo[b].size()) swap(a, b);
      printf("%d\n", ccount(a, b));
    }
  }
  return 0;
}
