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 long long N = 1e5 + 1, M = 2e5 + 1, inf = 1e18, mod = 1e9 + 7;
long long par[N], sz[N], ANS[N];
pair<long long, long long> p[N];
vector<pair<long long, long long> > cl[N];
vector<long long> clp[N], tmp;
bool is[N];
unordered_map<int, int> dg[N];
map<pair<long long, long long>, int> mp;
long long root(long long x) { return par[x] == x ? x : par[x] = root(par[x]); }
void mrg(long long x, long long y) {
  x = root(x);
  y = root(y);
  if (x != y) {
    if (sz[x] > sz[y]) swap(x, y);
    par[x] = y;
    sz[y] += sz[x];
    tmp.push_back(x);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  for (int i = 0; i < m; i++) {
    long long u, v, c;
    cin >> u >> v >> c;
    cl[c].push_back({u, v});
    dg[u][c]++;
    dg[v][c]++;
  }
  long long q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    long long u, v;
    cin >> u >> v;
    if (dg[u].size() > dg[v].size()) swap(u, v);
    p[i] = {u, v};
    if (mp.find(p[i]) != mp.end())
      is[i] = 1;
    else {
      mp[p[i]] = i;
      for (auto e : dg[u]) {
        clp[e.first].push_back(i);
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    for (auto e : cl[i]) mrg(e.first, e.second);
    for (auto e : clp[i])
      if (root(p[e].first) == root(p[e].second)) ANS[e]++;
    for (auto e : tmp) {
      par[e] = e;
      sz[e] = 1;
    }
    tmp.clear();
  }
  for (int i = 0; i < q; i++) {
    if (is[i])
      cout << ANS[mp[p[i]]] << "\n";
    else
      cout << ANS[i] << "\n";
  }
  return 0;
}
