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;
mt19937 rng(time(0));
const long long mod = 1e9 + 7;
const int mn = 2e5 + 10;
unordered_map<int, int> m[mn];
map<pair<int, int>, int> hail;
map<int, int> rn[mn];
int p[mn], r[mn];
int f(int x) { return x == p[x] ? x : (p[x] = f(p[x])); }
void mrg(int a, int b) {
  a = f(a), b = f(b);
  if (r[a] < r[b]) swap(a, b);
  if (r[a] == r[b]) r[a]++;
  p[b] = a;
}
int ct;
int main() {
  int n, k;
  iota(p, p + mn, 0);
  scanf("%d%d", &n, &k);
  for (int i = 0; i < k; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    int c1 = 0, c2 = 0;
    if (rn[a].count(c)) c1 = rn[a][c];
    if (rn[b].count(c)) c2 = rn[b][c];
    rn[a][c] = rn[b][c] = ++ct;
    if (c1) mrg(c1, ct);
    if (c2) mrg(c2, ct);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    if (rn[a].size() > rn[b].size()) swap(a, b);
    auto it = hail.find({a, b});
    if (it != hail.end()) {
      printf("%d\n", it->second);
      continue;
    }
    int ans = 0;
    for (pair<int, int> c : rn[a]) {
      if (rn[b].count(c.first) && f(rn[b][c.first]) == f(c.second)) ans++;
    }
    printf("%d\n", hail[{a, b}] = ans);
  }
}
