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;
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
double distance(int x1, int y1, int x2, int y2) {
  return sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2) * 1.0);
}
int factorial(int n) { return (n == 1 || n == 0) ? 1 : n * factorial(n - 1); }
bool comp(pair<int, double> a, pair<int, double> b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second > b.second;
}
const int X[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int Y[] = {0, 1, 0, -1, 1, -1, 1, -1};
vector<pair<int, int> > adj[101];
bool vis[105];
int sum = 0;
int c;
int ma[105];
bool dfs(int p, int h) {
  if (vis[p]) return 0;
  if (p == c) {
    return 1;
  }
  vis[p] = 1;
  if (h == 0) {
    for (auto i : adj[p]) {
      if (ma[i.second] != 1) {
        if (dfs(i.first, i.second)) {
          sum += 1;
          ma[i.second] = 1;
        }
      }
      for (int j = 0; j < 105; j++) vis[j] = 0;
    }
  } else {
    int u = 0;
    for (auto i : adj[p]) {
      if (i.second == h && vis[i.first] == 0) {
        if (dfs(i.first, i.second)) return 1;
        continue;
      } else
        continue;
    }
    return 0;
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
      int a, b, c1;
      cin >> a >> b >> c1;
      adj[a].push_back(make_pair(b, c1));
      adj[b].push_back(make_pair(a, c1));
    }
    int q;
    cin >> q;
    while (q--) {
      sum = 0;
      int a;
      cin >> a >> c;
      for (int i = 0; i < 105; i++) {
        vis[i] = 0;
        ma[i] = 0;
      }
      dfs(a, 0);
      cout << sum << '\n';
    }
  }
  return 0;
}
