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 maxn = 1e5 + 10;
const int sq = 500;
int n;
int grau[maxn];
int ind_heavy[maxn];
int pai[maxn], peso[maxn];
int ind_cc[maxn];
bool mark[maxn];
int ans[sq][maxn];
vector<int> comp[maxn];
vector<int> aparece[maxn];
vector<pair<int, int> > edge[maxn];
void init(void) {
  for (int i = 1; i <= n; i++) pai[i] = i, peso[i] = 1;
}
int Find(int x) {
  if (pai[x] == x) return x;
  return pai[x] = Find(pai[x]);
}
void Join(int x, int y) {
  x = Find(x), y = Find(y);
  if (x == y) return;
  if (peso[x] < peso[y]) swap(x, y);
  pai[y] = x, peso[x] += peso[y];
}
int main(void) {
  int m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v, c;
    scanf("%d %d %d", &u, &v, &c);
    edge[c].push_back({u, v});
    grau[u]++, grau[v]++;
  }
  init();
  int cc = 0;
  for (int i = 1; i <= m; i++) {
    for (auto e : edge[i]) {
      int u = e.first, v = e.second;
      Join(u, v);
    }
    for (auto e : edge[i]) {
      int u = e.first, v = e.second;
      if (Find(u) == u && ind_cc[u] == 0) ind_cc[u] = ++cc;
      if (Find(v) == v && ind_cc[v] == 0) ind_cc[v] = ++cc;
    }
    for (auto e : edge[i]) {
      int u = e.first, v = e.second;
      if (mark[u] == 0) {
        comp[ind_cc[Find(u)]].push_back(u);
        aparece[u].push_back(ind_cc[Find(u)]);
        mark[u] = 1;
      }
      if (mark[v] == 0) {
        comp[ind_cc[Find(v)]].push_back(v);
        aparece[v].push_back(ind_cc[Find(v)]);
        mark[v] = 1;
      }
    }
    for (auto e : edge[i]) {
      int u = e.first, v = e.second;
      pai[u] = u, pai[v] = v;
      peso[u] = peso[v] = 1;
      ind_cc[u] = ind_cc[v] = mark[u] = mark[v] = 0;
    }
  }
  int ind_h = 0;
  for (int i = 1; i <= n; i++)
    if (grau[i] > sq) ind_heavy[i] = ++ind_h;
  for (int i = 1; i <= n; i++)
    if (ind_heavy[i] != 0)
      for (auto j : aparece[i])
        for (auto v : comp[j]) ans[ind_heavy[i]][v]++;
  for (int i = 1; i <= n; i++) sort(aparece[i].begin(), aparece[i].end());
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    if (ind_heavy[u]) {
      printf("%d\n", ans[ind_heavy[u]][v]);
      continue;
    }
    if (ind_heavy[v]) {
      printf("%d\n", ans[ind_heavy[v]][u]);
      continue;
    }
    int l = 0, r = 0;
    int tot = 0;
    while (l < aparece[u].size() && r < aparece[v].size()) {
      if (aparece[u][l] == aparece[v][r]) {
        tot++;
        l++, r++;
      } else if (aparece[u][l] < aparece[v][r])
        l++;
      else
        r++;
    }
    printf("%d\n", tot);
  }
}
