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 N = (int)1e5 + 5;
const int M = 600;
vector<pair<int, int> > col[N];
vector<int> big;
vector<int> par[N], sz[N];
map<int, int> reName[N];
map<pair<int, int>, int> small;
int n, m;
int a[2 * N];
int s;
void readEdge() {
  int v, u;
  int c;
  scanf("%d%d%d", &v, &u, &c);
  col[c].push_back(make_pair(v, u));
  return;
}
void read() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) readEdge();
  return;
}
void incSmall(int v, int u) {
  pair<int, int> tmp = make_pair(v, u);
  if (small.count(tmp) == 0)
    small[tmp] = 1;
  else
    small[tmp]++;
  return;
}
int getPar(int c, int v) {
  return (par[c][v] == -1 ? v : par[c][v] = getPar(c, par[c][v]));
}
void unite(int c, int v, int u) {
  v = getPar(c, v);
  u = getPar(c, u);
  if (v == u) return;
  if (sz[c][v] < sz[c][u]) swap(v, u);
  sz[c][v] += sz[c][u];
  par[c][u] = v;
  return;
}
void solveColor(int c) {
  if ((int)col[c].size() == 0) return;
  int esz = (int)col[c].size();
  s = 0;
  for (int i = 0; i < esz; i++) {
    a[s++] = col[c][i].first;
    a[s++] = col[c][i].second;
  }
  sort(a, a + s);
  s = unique(a, a + s) - a;
  for (int i = 0; i < s; i++) reName[c][a[i]] = i;
  par[c].resize(s);
  sz[c].resize(s);
  for (int i = 0; i < s; i++) {
    par[c][i] = -1;
    sz[c][i] = 1;
  }
  for (int i = 0; i < esz; i++)
    unite(c, reName[c][col[c][i].first], reName[c][col[c][i].second]);
  if (s > M)
    big.push_back(c);
  else {
    for (int i = 0; i < s; i++)
      for (int j = i + 1; j < s; j++)
        if (getPar(c, i) == getPar(c, j)) incSmall(a[i], a[j]);
  }
  return;
}
void solve() {
  for (int i = 1; i <= m; i++) solveColor(i);
  return;
}
void query() {
  int v, u;
  scanf("%d%d", &v, &u);
  if (v > u) swap(v, u);
  int ans = small[make_pair(v, u)];
  for (int i = 0; i < (int)big.size(); i++) {
    int c = big[i];
    if (reName[c].count(v) == 0 || reName[c].count(u) == 0) continue;
    ans += (int)(getPar(c, reName[c][v]) == getPar(c, reName[c][u]));
  }
  printf("%d\n", ans);
  return;
}
int main() {
  read();
  solve();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) query();
  return 0;
}
