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;
long long const N = 1e5 + 1, mod = 1e9 + 7, inf = 1e17;
int par[N], sz[N], u[N], v[N], ans[N];
vector<pair<int, int> > color[N];
vector<int> vec, col[N];
unordered_map<int, int> mp, vt[N];
map<pair<int, int>, int> anss;
bool mark[N];
long long get(long long a) { return par[a] == a ? a : par[a] = get(par[a]); }
void add(long long a, long long b) {
  long long a1 = get(a), a2 = get(b);
  if (a1 != a2) {
    if (sz[a1] > sz[a2]) swap(a1, a2);
    par[a1] = a2;
    sz[a2] += sz[a1];
    vec.push_back(a1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  for (int i = 0; i < m; i++) {
    long long u, v, cl;
    cin >> u >> v >> cl;
    color[cl].push_back({u, v});
    vt[u][cl]++;
    vt[v][cl]++;
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> u[i] >> v[i];
    if (vt[u[i]].size() > vt[v[i]].size()) swap(u[i], v[i]);
    if (anss.find({u[i], v[i]}) == anss.end()) {
      anss[{u[i], v[i]}] = i;
      for (auto x : vt[u[i]]) {
        col[x.first].push_back(i);
      }
    } else
      mark[i] = 1;
  }
  for (int i = 1; i <= m; i++) {
    for (auto x : color[i]) {
      add(x.first, x.second);
    }
    for (auto x : col[i]) {
      if (get(u[x]) == get(v[x])) ans[x]++;
    }
    for (auto x : vec) {
      par[x] = x;
      sz[x] = 1;
      sz[par[x]] = 1;
    }
    vec.clear();
  }
  for (int i = 0; i < q; i++) {
    if (mark[i])
      cout << ans[anss[{u[i], v[i]}]] << "\n";
    else
      cout << ans[i] << "\n";
  }
}
