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;
template <typename R, typename D>
inline void Min(R &a, D b) {
  if (a > b) a = b;
}
template <typename D, typename R>
inline void Max(D &a, R b) {
  if (a < b) a = b;
}
const int N = 100005;
int n, m, goc[N + N], sdeg[N + N];
vector<int> a[N];
map<pair<int, int>, int> dd;
pair<pair<int, int>, int> q[N];
int root(int u) { return goc[u] == 0 ? u : goc[u] = root(goc[u]); }
void Weave(int u, int v) {
  u = root(u);
  v = root(v);
  if (u == v) return;
  goc[u] = v;
}
void In() {
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> q[i].first.first >> q[i].first.second >> q[i].second;
    a[q[i].first.first].push_back(q[i].second);
    a[q[i].first.second].push_back(q[i].second);
  }
}
int getpos(const vector<int> &ve, int u, int c) {
  return sdeg[u - 1] + lower_bound(ve.begin(), ve.end(), c) - ve.begin();
}
void Pre() {
  sdeg[0] = 1;
  for (int i = 1; i <= n; ++i) {
    sort(a[i].begin(), a[i].end());
    a[i].resize(unique(a[i].begin(), a[i].end()) - a[i].begin());
    sdeg[i] = sdeg[i - 1] + a[i].size();
  }
  for (int i = 1; i <= m; ++i) {
    int u = q[i].first.first;
    int v = q[i].first.second;
    int c = q[i].second;
    Weave(getpos(a[u], u, c), getpos(a[v], v, c));
  }
}
void Out() {
  int T;
  cin >> T;
  while (T--) {
    int u, v;
    int cnt = 0;
    cin >> u >> v;
    if (a[u].size() == a[v].size()) {
      if (u > v) swap(u, v);
    } else if (a[u].size() > a[v].size())
      swap(u, v);
    if (dd.count(make_pair(u, v))) {
      cout << dd[make_pair(u, v)] << '\n';
      continue;
    }
    for (int i = 0; i < a[u].size(); ++i) {
      int temp = a[u][i];
      if (binary_search(a[v].begin(), a[v].end(), temp)) {
        if (root(sdeg[u - 1] + i) == root(getpos(a[v], v, temp))) ++cnt;
      }
    }
    dd[make_pair(u, v)] = cnt;
    cout << cnt << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  In();
  Pre();
  Out();
  return 0;
}
