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 power(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
string a, b, c;
int q, n, t;
int arr[100009];
int const mi = 1e5 + 9;
unordered_map<int, int> par[mi];
int root(int v, int c) {
  return par[v][c] < 0 ? v : (par[v][c] = root(par[v][c], c));
}
void merge(int x, int y, int c) {
  if ((x = root(x, c)) == (y = root(y, c))) return;
  if (par[y][c] < par[x][c]) swap(x, y);
  par[x][c] += par[y][c];
  par[y][c] = x;
}
int brr[100009];
int m;
set<int> s[int(1e5 + 4)];
map<pair<int, int>, int> seen;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    if (par[y].find(c) == par[y].end()) par[y][c] = -1;
    if (par[x].find(c) == par[x].end()) par[x][c] = -1;
    merge(x, y, c);
  }
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    int an = 0;
    if (par[x].size() > par[y].size()) swap(x, y);
    if (seen.find(make_pair(x, y)) != seen.end()) {
      cout << seen[make_pair(x, y)] << "\n";
      continue;
    }
    for (auto &ig : par[x]) {
      int i = ig.first;
      if (par[y].find(i) == par[y].end()) continue;
      if (root(x, i) == root(y, i)) an++;
    }
    seen[make_pair(x, y)] = an;
    cout << an << "\n";
  }
}
