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 rec {
  int x, y, co;
  void read() { scanf("%d%d%d", &x, &y, &co); }
} E[200005];
struct rzz {
  int id, co;
} opt[200005], R[200005];
bool cmp(const rec &i, const rec &j) { return i.co < j.co; }
bool cm1(const rzz &i, const rzz &j) { return i.co < j.co; }
bool use[200005];
int fa[200005], stk[200005], top, cnt, tot;
int n, m, Q, Time;
map<pair<int, int>, int> S;
map<int, bool> orz[200005];
int getfa(int v) {
  if (fa[v] == v) return v;
  fa[v] = getfa(fa[v]);
  return fa[v];
}
void link(int x, int y) {
  if (!use[x]) stk[++top] = x;
  use[x] = 1;
  if (!use[y]) stk[++top] = y;
  use[y] = 1;
  int fx = getfa(x), fy = getfa(y);
  if (fx != fy) fa[fx] = fy;
}
void doit() {
  for (int k = 1; k <= top; k++) {
    int i = stk[k];
    R[k] = (rzz){i, getfa(i)};
  }
  sort(R + 1, R + 1 + top, cm1);
  for (int i = 1, j = 1; i <= top; i = j) {
    for (; j <= top && R[i].co == R[j].co; j++)
      ;
    if (j - i > 200) {
      ++cnt;
      for (int k = i; k <= j - 1; k++) orz[cnt][R[k].id] = 1;
    } else {
      for (int k = i; k <= j - 1; k++)
        for (int l = i; l <= j - 1; l++) {
          S[pair<int, int>(R[k].id, R[l].id)]++;
          tot++;
        }
    }
  }
  for (int k = 1; k <= top; k++) {
    int i = stk[k];
    fa[i] = i;
    use[i] = 0;
  }
  top = 0;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= m; i++) E[i].read();
  sort(E + 1, E + 1 + m, cmp);
  for (int i = 1, j = 1; i <= m; i = j) {
    for (; j <= m && E[i].co == E[j].co; j++) link(E[j].x, E[j].y);
    doit();
  }
  scanf("%d", &Q);
  while (Q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    int ans = S[pair<int, int>(x, y)];
    for (int i = 1; i <= cnt; i++)
      if (orz[i][x] && orz[i][y]) ans++;
    printf("%d\n", ans);
  }
}
