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 N = 100010;
int par_uf[N], rank_uf[N];
void init(int n) {
  for (int i = 0; i < n; i++) {
    par_uf[i] = i;
    rank_uf[i] = 0;
  }
}
int find(int x) {
  if (par_uf[x] == x)
    return x;
  else
    return par_uf[x] = find(par_uf[x]);
}
void unite(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  if (rank_uf[x] < rank_uf[y])
    par_uf[x] = y;
  else {
    par_uf[y] = x;
    if (rank_uf[x] == rank_uf[y]) rank_uf[x]++;
  }
}
bool same(int x, int y) { return find(x) == find(y); }
int n, m, q, qa[N], qb[N];
vector<pair<int, int> > edge[N];
vector<vector<int> > comp;
vector<int> appear[N];
const int B = 450;
vector<int> big;
int bigID[N];
int table[B + 10][N];
int used[N];
vector<int> temp[N];
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 0; i < (m); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    c--;
    edge[c].push_back(make_pair(a, b));
  }
  cin >> q;
  for (int i = 0; i < (q); i++) {
    cin >> qa[i] >> qb[i];
    qa[i]--;
    qb[i]--;
  }
  init(n);
  for (int c = 0; c < (N); c++) {
    for (int i = 0; i < (((int)(edge[c]).size())); i++) {
      int a = edge[c][i].first, b = edge[c][i].second;
      unite(a, b);
    }
    for (int i = 0; i < (((int)(edge[c]).size())); i++) {
      int a = edge[c][i].first, b = edge[c][i].second;
      if (!used[a]) {
        temp[find(a)].push_back(a);
        used[a] = 1;
      }
      if (!used[b]) {
        temp[find(b)].push_back(b);
        used[b] = 1;
      }
    }
    for (int i = 0; i < (((int)(edge[c]).size())); i++) {
      int a = edge[c][i].first, b = edge[c][i].second;
      if (((int)(temp[a]).size()) > 0) {
        comp.push_back(temp[a]);
        temp[a].clear();
      }
      if (((int)(temp[b]).size()) > 0) {
        comp.push_back(temp[b]);
        temp[b].clear();
      }
      used[a] = 0;
      used[b] = 0;
      par_uf[a] = a;
      par_uf[b] = b;
      rank_uf[a] = 0;
      rank_uf[b] = 0;
    }
  }
  for (int i = 0; i < (((int)(comp).size())); i++) {
    for (int j = 0; j < (((int)(comp[i]).size())); j++) {
      appear[comp[i][j]].push_back(i);
    }
  }
  for (int i = 0; i < (n); i++) {
    if (((int)(appear[i]).size()) > B) {
      bigID[i] = ((int)(big).size());
      big.push_back(i);
    } else {
      bigID[i] = -1;
    }
  }
  for (int i = 0; i < (((int)(big).size())); i++) {
    int b = big[i];
    for (int j = 0; j < (((int)(appear[b]).size())); j++) {
      int c = appear[b][j];
      for (int k = 0; k < (((int)(comp[c]).size())); k++) {
        table[i][comp[c][k]]++;
      }
    }
  }
  for (int _ = 0; _ < (q); _++) {
    int a = qa[_], b = qb[_];
    if (bigID[a] == -1) {
      swap(a, b);
    }
    if (bigID[a] != -1) {
      cout << table[bigID[a]][b] << endl;
    } else {
      int cur1 = 0, cur2 = 0, res = 0;
      while (cur1 < ((int)(appear[a]).size()) &&
             cur2 < ((int)(appear[b]).size())) {
        if (appear[a][cur1] == appear[b][cur2]) {
          res++;
          cur1++;
          cur2++;
        } else if (appear[a][cur1] < appear[b][cur2]) {
          cur1++;
        } else {
          cur2++;
        }
      }
      cout << res << endl;
    }
  }
}
