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 = 2e5 + 10;
int par[MAXN], size[MAXN];
int find(int i) {
  if (par[i] != i) par[i] = find(par[i]);
  return par[i];
}
void join(int a, int b) {
  int x = find(a), y = find(b);
  if (x != y) {
    if (size[x] >= size[y])
      size[x] += size[y], par[y] = x;
    else
      size[y] += size[x], par[x] = y;
  }
}
vector<int> s[MAXN];
vector<pair<int, int> > g[MAXN];
map<pair<int, int>, int> done;
int sum[MAXN];
int get(int pos, int c) {
  int id = lower_bound(s[pos].begin(), s[pos].end(), c) - s[pos].begin();
  if (id < ((int)(s[pos]).size()) && s[pos][id] == c) return sum[pos] + id;
  return -1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, x, y, c, t = 1, q;
  cin >> n >> m;
  for (int i = 0, ThxDem = MAXN; i < ThxDem; ++i) par[i] = i, size[i] = 1;
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) {
    cin >> x >> y >> c;
    x--;
    y--;
    s[x].push_back(c);
    s[y].push_back(c);
    g[x].push_back({y, c});
    g[y].push_back({x, c});
  }
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) {
    sort(s[i].begin(), s[i].end());
    s[i].erase(unique(s[i].begin(), s[i].end()), s[i].end());
    sum[i + 1] = sum[i] + ((int)(s[i]).size());
  }
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) {
    for (auto x : g[i]) {
      join(get(i, x.second), get(x.first, x.second));
    }
  }
  cin >> q;
  while (q--) {
    cin >> x >> y;
    x--;
    y--;
    if (((int)(g[x]).size()) > ((int)(g[y]).size())) swap(x, y);
    if (done.count({x, y}))
      cout << done[{x, y}] << "\n";
    else {
      int ans = 0;
      for (int i = 0, ThxDem = ((int)(s[x]).size()); i < ThxDem; ++i) {
        int aux = get(y, s[x][i]);
        ans += aux >= 0 && find(sum[x] + i) == find(aux);
      }
      done[{x, y}] = ans;
      cout << ans << "\n";
    }
  }
}
