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;
using LL = int64_t;
const int INF = 0x3fffffff;
const double EPS = 1e-9;
const double PI = 2 * asin(1);
struct Edge {
  int v, c;
  Edge(int v = 0, int c = 0) : v(v), c(c) {}
};
int g_n, g_m, g_u, g_v;
vector<bool> vb_mark;
vector<vector<Edge>> vve;
vector<vector<int>> vvi_ans;
void pretreat() {}
bool input() {
  cin >> g_n >> g_m;
  if (cin.eof()) return false;
  vve.clear(), vb_mark.clear(), vvi_ans.clear();
  vb_mark.resize(g_n + 1), vve.resize(g_n + 1);
  vvi_ans.resize(g_n + 1);
  for (int i = 0; i <= g_n; ++i) vvi_ans[i].resize(g_n + 1);
  int u, v, c;
  for (int i = 0; i < g_m; ++i) {
    scanf("%d%d%d", &u, &v, &c);
    vve[u].emplace_back(v, c);
    vve[v].emplace_back(u, c);
  }
  return true;
}
void dfs(int u, int c, set<int> &sti) {
  vb_mark[u] = true;
  for (Edge &e : vve[u]) {
    if (!vb_mark[e.v] && e.c == c) {
      if (e.v == g_v) {
        sti.insert(c);
        return;
      }
      vb_mark[e.v] = true;
      dfs(e.v, c, sti);
    }
  }
}
void solve() {
  int k;
  cin >> k;
  while (k--) {
    scanf("%d%d", &g_u, &g_v);
    if (vvi_ans[g_u][g_v] > 0)
      printf("%d\n", vvi_ans[g_u][g_v]);
    else if (vvi_ans[g_u][g_v] < 0)
      printf("%d\n", 0);
    else {
      set<int> sti;
      for (Edge &e : vve[g_u]) {
        if (e.v == g_v) {
          sti.insert(e.c);
          continue;
        }
        fill(vb_mark.begin(), vb_mark.end(), false);
        vb_mark[g_u] = true;
        dfs(e.v, e.c, sti);
      }
      vvi_ans[g_u][g_v] = vvi_ans[g_v][g_u] = sti.size();
      printf("%d\n", vvi_ans[g_u][g_v]);
    }
  }
}
int main() {
  pretreat();
  while (input()) {
    solve();
  }
  return 0;
}
