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;
const int MAXN = (int)((1e5) + 10);
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int max(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
int min(int a, int b) {
  if (a < b)
    return a;
  else
    return b;
}
bool isPrime(int N) {
  for (int i = 2; i * i <= N; ++i) {
    if (N % i == 0) return false;
  }
  return true;
}
int cbrt(int x) {
  int lo = 1, hi = min(2000000ll, x);
  while (hi - lo > 1) {
    int mid = (lo + hi) / 2;
    if (mid * mid * mid < x) {
      lo = mid;
    } else
      hi = mid;
  }
  if (hi * hi * hi <= x)
    return hi;
  else
    return lo;
}
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
const int nax = (int)(100000 + 10);
vector<string> v;
vector<int> vec[27], adj[1100][1100];
stack<int> st;
string ss;
long long int a, b, c, pre = 0, n, m, k = 0, f = 0, ans = 0, h = 0, t, res = 0;
long long int dp[301], act[305002], vis[110][110];
map<int, map<int, int>> ab;
void dfs(int u, int col) {
  vis[u][col] = 1;
  if (u == b) {
    h = 1;
    return;
  }
  for (auto &i : adj[u][col]) {
    if (!vis[i][col]) {
      dfs(i, col);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    adj[a][c].push_back(b);
    adj[b][c].push_back(a);
  }
  cin >> res;
  while (res--) {
    cin >> a >> b;
    for (int i = 1; i <= 100; i++) {
      h = 0;
      dfs(a, i);
      ans += h;
    }
    for (int i = 0; i < 105; i++) {
      for (int j = 0; j < 105; j++) {
        vis[i][j] = 0;
      }
    }
    cout << ans << "\n";
    ans = 0;
  }
  return 0;
}
