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 * 1000 + 10;
struct dsu {
  set<pair<int, int> > node;
  vector<int> par;
  vector<int> sz;
  int cnt = 0;
  void add(int v) {
    if (node.lower_bound(pair<int, int>(v, 0)) != node.end() &&
        node.lower_bound(pair<int, int>(v, 0))->first == v)
      return;
    node.insert(pair<int, int>(v, cnt++));
  }
  void merge(int u, int v) {
    u = node.lower_bound(pair<int, int>(u, 0))->second;
    v = node.lower_bound(pair<int, int>(v, 0))->second;
    u = find_par(u);
    v = find_par(v);
    if (u == v) return;
    if (sz[u] > sz[v]) swap(u, v);
    par[u] = v;
    sz[v] += sz[u];
  }
  int find_par(int u) {
    return (par[u] == -1 ? u : (par[u] = find_par(par[u])));
  }
  void prep() {
    for (int i = 0; i < node.size(); i++) {
      sz.push_back(0);
      par.push_back(-1);
    }
  }
} g[maxn];
int n, m, q;
vector<pair<int, int> > col[maxn];
vector<pair<int, int> > adj[maxn];
map<pair<int, int>, int> mp;
vector<int> cl[maxn], ans;
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    u--, v--, c--;
    adj[u].push_back(pair<int, int>(u, c));
    adj[v].push_back(pair<int, int>(v, c));
    col[c].push_back(pair<int, int>(u, v));
    g[c].add(u);
    g[c].add(v);
    cl[u].push_back(c);
    cl[v].push_back(c);
  }
  for (int i = 0; i < n; i++) {
    if (!cl[i].size()) continue;
    sort(cl[i].begin(), cl[i].end());
    cl[i].resize(unique(cl[i].begin(), cl[i].end()) - cl[i].begin());
  }
  for (int i = 0; i < maxn; i++) g[i].prep();
  for (int i = 0; i < maxn; i++)
    for (pair<int, int> tmp : col[i]) g[i].merge(tmp.first, tmp.second);
  cin >> q;
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    if (u > v) swap(u, v);
    if (mp.count(pair<int, int>(u, v))) {
      ans.push_back(mp[pair<int, int>(u, v)]);
      continue;
    }
    if (cl[u].size() > cl[v].size()) swap(u, v);
    int res = 0;
    for (int c : cl[u]) {
      int ind = lower_bound(cl[v].begin(), cl[v].end(), c) - cl[v].begin();
      if (ind == cl[v].size() || cl[v][ind] != c) continue;
      int u2 = g[c].node.lower_bound(pair<int, int>(u, 0))->second,
          v2 = g[c].node.lower_bound(pair<int, int>(v, 0))->second;
      u2 = g[c].find_par(u2), v2 = g[c].find_par(v2);
      if (u2 == v2) res++;
    }
    mp[pair<int, int>(min(u, v), max(u, v))] = res;
    ans.push_back(res);
  }
  for (int op : ans) cout << op << '\n';
  return 0;
}
