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;
struct Edge {
  int x, y, z;
} e[100005];
int n, ne, idx = 0;
int trc[200005], sz[100005], d[200005];
map<int, int> v[100005];
map<pair<int, int>, int> mymap;
int root(int i) { return i == trc[i] ? i : trc[i] = root(trc[i]); }
void Union(int x, int y) {
  x = root(x), y = root(y);
  if (x == y) return;
  if (d[x] < d[y])
    d[y] += d[x], trc[x] = y;
  else
    d[x] += d[y], trc[y] = x;
}
void solve() {
  int i, nq, x, y, res, id1, id2, color;
  map<int, int>::iterator ite;
  for ((i) = (1); (i) <= (200000); (i)++) trc[i] = i, d[i] = 1;
  for ((i) = 0; (i) < (ne); (i)++) {
    if (v[e[i].x].find(e[i].z) == v[e[i].x].end()) {
      v[e[i].x][e[i].z] = ++idx;
      sz[e[i].x]++;
    }
    if (v[e[i].y].find(e[i].z) == v[e[i].y].end()) {
      v[e[i].y][e[i].z] = ++idx;
      sz[e[i].y]++;
    }
    Union(v[e[i].x][e[i].z], v[e[i].y][e[i].z]);
  }
  scanf("%d", &nq);
  while (nq--) {
    scanf("%d %d", &x, &y);
    if (x > y) (x) ^= (y), (y) ^= (x), (x) ^= (y);
    if (mymap.find(make_pair(x, y)) != mymap.end()) {
      printf("%d\n", mymap[make_pair(x, y)]);
      continue;
    }
    if (sz[x] > sz[y]) (x) ^= (y), (y) ^= (x), (x) ^= (y);
    res = 0;
    for (ite = v[x].begin(); ite != v[x].end(); ite++) {
      color = ite->first;
      id1 = ite->second;
      if (v[y].find(color) != v[y].end()) {
        id2 = v[y][color];
        if (root(id1) == root(id2)) res++;
      }
    }
    printf("%d\n", res);
    if (x > y) (x) ^= (y), (y) ^= (x), (x) ^= (y);
    mymap[make_pair(x, y)] = res;
  }
}
void input() {
  int i;
  scanf("%d %d", &n, &ne);
  for ((i) = 0; (i) < (ne); (i)++) scanf("%d %d %d", &e[i].x, &e[i].y, &e[i].z);
  solve();
}
int main() {
  input();
  return 0;
}
