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 mv4[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int mv8[8][2] = {{0, 1}, {0, -1}, {1, 0},  {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
const int inf = 0x7fffffff;
const int _inf = 0x8fffffff;
const long long INF = 1e18;
const double EPS = 1e-8;
const double pi = acos(-1.0);
double ROUNDF(double x, int n) {
  double t1, t2 = modf(x, &t1);
  double t3 = pow(10, n);
  long t4 = long(t2 * t3 + 0.5);
  x = t1 + t4 / t3;
  return x;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int maxn = 100 + 5;
vector<int> pic[maxn][maxn];
int n, m, ans;
int vis[maxn];
void dfs(int beg, int end, int col) {
  vis[beg] = 1;
  if (beg == end) {
    ans++;
    return;
  }
  if (pic[beg][col].size()) {
    for (int(i) = (0); (i) <= (pic[beg][col].size() - 1); (i)++) {
      int v = pic[beg][col][i];
      if (!vis[v]) dfs(v, end, col);
    }
  }
}
int main() {
  ;
  scanf("%d%d", &n, &m);
  for (int(i) = (1); (i) <= (m); (i)++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    pic[a][c].push_back(b);
    pic[b][c].push_back(a);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b;
    ans = 0;
    scanf("%d%d", &a, &b);
    for (int(i) = (1); (i) <= (m); (i)++) {
      memset((vis), (0), sizeof(vis));
      dfs(a, b, i);
    }
    printf("%d\n", ans);
  }
  return 0;
}
