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 dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int ddx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, ddy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
void mad(long long &a, long long b) {
  a = (a + b) % 1000000007;
  if (a < 0) a += 1000000007;
}
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int n, m;
unordered_map<int, int> pa[100005];
unordered_map<int, int> sz[100005];
vector<pair<int, int> > ed[100005];
set<int> cols[100005];
int find(int u, int c) {
  if (pa[c].find(u) == pa[c].end()) return 0;
  if (u == pa[c][u]) return u;
  return pa[c][u] = find(pa[c][u], c);
}
void join(int u, int v, int c) {
  int a = find(u, c);
  int b = find(v, c);
  if (sz[c][a] < sz[c][b]) swap(a, b);
  pa[c][b] = pa[c][a];
  sz[c][a] += sz[c][b];
  sz[c][b] = 0;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    ed[c].push_back(pair<int, int>(u, v));
    cols[u].insert(c);
    cols[v].insert(c);
  }
  for (int c = 1; c <= m; c++) {
    for (pair<int, int> e : ed[c]) {
      pa[c][e.first] = e.first;
      pa[c][e.second] = e.second;
    }
    for (pair<int, int> e : ed[c]) {
      if (find(e.first, c) == find(e.second, c)) continue;
      join(e.first, e.second, c);
    }
    for (pair<int, int> e : ed[c]) {
      find(e.first, c);
      find(e.second, c);
    }
  }
  int q;
  cin >> q;
  unordered_map<int, unordered_map<int, int> > res;
  while (q--) {
    int u, v;
    cin >> u >> v;
    if (cols[u].size() > cols[v].size()) swap(u, v);
    if (res.find(u) != res.end() && res[u].find(v) != res[u].end()) {
      cout << res[u][v] << "\n";
      continue;
    }
    int ans = 0;
    for (int c : cols[u]) {
      if (cols[v].find(c) == cols[v].end()) continue;
      if (pa[c][u] == pa[c][v]) ans++;
    }
    res[u][v] = ans;
    cout << ans << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}
