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 = 100 + 10;
const int maxe = maxn * 2;
int fir[maxn];
struct Edge {
  int v, c, nxt;
  Edge(int v = 0, int c = 0, int nxt = 0) : v(v), c(c), nxt(nxt) {}
} edge[maxe];
int cnt;
bool cir[maxn];
void init() {
  memset(fir, -1, sizeof(fir));
  cnt = 0;
}
void add_edge(int u, int v, int c) {
  edge[cnt] = Edge(v, c, fir[u]);
  fir[u] = cnt++;
}
bool dfs(int from, int to, int c) {
  cir[from] = true;
  if (from == to) return true;
  for (int e = fir[from]; e != -1; e = edge[e].nxt)
    if (!cir[edge[e].v]) {
      if (edge[e].c != c) continue;
      if (dfs(edge[e].v, to, c)) {
        return true;
      }
    }
  return false;
}
int solve(int from, int to) {
  int ans = 0;
  bool vis[maxn];
  memset(vis, false, sizeof(vis));
  for (int e = fir[from]; e != -1; e = edge[e].nxt)
    if (!vis[edge[e].c]) {
      memset(cir, false, sizeof(cir));
      if (dfs(from, to, edge[e].c)) {
        ans++;
      }
      vis[edge[e].c] = true;
    }
  return ans;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  init();
  for (int i = 0; i < m; i++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    add_edge(u, v, c);
    add_edge(v, u, c);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    printf("%d\n", solve(u, v));
  }
  return 0;
}
