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;
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long N = 1e5 + 5;
const long long M = 5e5 + 5;
const long long L = 2e1 + 1;
const long long inf = 9e17;
const long long mod = 1e9 + 7;
long long xyz = 1;
long long n, m, q;
long long deg[N];
unordered_map<long long, long long> mem[N];
unordered_map<long long, long long> ids[N];
vector<long long> con[N];
long long fnd(long long i, long long c) {
  return ids[i].count(c) ? ids[i][c] = fnd(ids[i][c], c) : i;
}
void unn(long long a, long long b, long long c) {
  a = fnd(a, c);
  b = fnd(b, c);
  if (a != b) ids[a][c] = b;
}
void run() {
  cin >> n >> m;
  for (long long i = 1; i <= m; i++) {
    long long u;
    long long v;
    long long c;
    cin >> u >> v >> c;
    deg[u]++;
    deg[v]++;
    con[u].push_back(c);
    con[v].push_back(c);
    unn(u, v, c);
  }
  for (long long i = 1; i <= n; i++)
    sort(begin(con[i]), end(con[i])),
        con[i].erase(unique(begin(con[i]), end(con[i])), con[i].end());
  cin >> q;
  while (q--) {
    long long u;
    long long v;
    cin >> u >> v;
    if (u > v) swap(u, v);
    if (mem[u].count(v)) {
      cout << mem[u][v] << "\n";
      continue;
    }
    long long x = deg[u] < deg[v] ? u : v;
    long long res = 0;
    for (long long c : con[x])
      if (fnd(u, c) == fnd(v, c)) res++;
    cout << (mem[u][v] = res) << "\n";
  }
}
signed main() {
  setIO();
  while (xyz--) run();
  return 0;
}
