Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
const int mod = 1e9 + 7;
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first < b.first;
}
long long exp(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b % 2) ans = ans * a;
    a = a * a;
    b /= 2;
  }
  return ans;
}
bool has[105];
long long comp[105];
bool vis[105];
vector<pair<long long, long long> > v[105];
void dfs(long long st, long long cmp) {
  vis[st] = 1;
  comp[st] = cmp;
  for (long long i = (long long)0; i < (long long)v[st].size(); i++) {
    long long to = v[st][i].first;
    if (!vis[to]) {
      dfs(to, cmp);
    }
  }
}
void dfs1(long long st, long long co) {
  vis[st] = 1;
  for (long long i = (long long)0; i < (long long)v[st].size(); i++) {
    long long to = v[st][i].first;
    long long col = v[st][i].second;
    if (!vis[to] && co == col) {
      dfs1(to, co);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = (long long)0; i < (long long)m; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    v[a].push_back({b, c});
    v[b].push_back({a, c});
    has[c] = 1;
  }
  for (long long i = (long long)1; i < (long long)n + 1; i++) {
    if (!vis[i]) dfs(i, i);
  }
  long long q;
  cin >> q;
  for (long long i = (long long)0; i < (long long)q; i++) {
    long long a, b;
    cin >> a >> b;
    if (comp[a] == comp[b]) {
      long long cnt = 0;
      for (long long i = (long long)1; i < (long long)105; i++) {
        if (has[i]) {
          memset((vis), 0, sizeof((vis)));
          dfs1(a, i);
          if (vis[b]) cnt++;
        }
      }
      cout << cnt << endl;
    }
  }
  return 0;
}
