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;
inline bool isEq(double a, double b) { return abs(a - b) < 1e-9; }
const double PI = acos(-1.0);
int relation[105][105];
int findRep(int c, int n) {
  if (n == relation[c][n])
    return n;
  else
    return relation[c][n] = findRep(c, relation[c][n]);
}
void makeUnion(int color, int x, int y) {
  int a = findRep(color, x);
  int b = findRep(color, y);
  if (a != b) relation[color][a] = b;
}
bool isSame(int color, int x, int y) {
  return findRep(color, x) == findRep(color, y);
}
vector<int> cols[105];
int count(int x, int y) {
  bitset<105> bs;
  bs.reset();
  if (cols[x].size() > cols[y].size()) swap(x, y);
  int d = cols[x].size();
  int cnt = 0;
  for (int i = 0; i < d; i++) {
    if (!bs[cols[x][i]]) {
      cnt += isSame(cols[x][i], x, y);
      bs[cols[x][i]] = 1;
    }
  }
  return cnt;
}
int main(int argc, char* argv[]) {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int j = 1; j <= m; j++)
    for (int i = 1; i <= n; i++) relation[j][i] = i;
  int a, b, c;
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &a, &b, &c);
    makeUnion(c, a, b);
    cols[a].push_back(c);
    cols[b].push_back(c);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &a, &b);
    printf("%d\n", count(a, b));
  }
  return 0;
}
