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;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcountll(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int n, m;
map<pair<int, int>, int> f;
map<pair<int, int>, vector<int> > V;
map<pair<int, int>, bool> have;
int coun = 0;
int MAX = 300;
bool d[100005][350];
int from[100005], to[100005], cost[100005], que[100005], size;
void go(int u, int c) {
  size = 0;
  que[size++] = u;
  have[make_pair(u, c)] = 1;
  for (int i = 0; i < (size); ++i) {
    u = que[i];
    pair<int, int> P = make_pair(u, c);
    for (int j = 0; j < (((int)(V[P]).size())); ++j) {
      int v = V[P][j];
      if (!have[make_pair(v, c)]) {
        have[make_pair(v, c)] = 1;
        que[size++] = v;
      }
    }
  }
  if (size <= MAX) {
    for (int i = 0; i < (size); ++i)
      for (int j = i + 1; j < size; j++) {
        int u = que[i], v = que[j];
        if (u > v) swap(u, v);
        f[make_pair(u, v)]++;
      }
  } else {
    coun++;
    for (int i = 0; i < (size); ++i) {
      int u = que[i];
      d[u][coun] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  int u, v, c;
  memset(d, 0, sizeof(d));
  for (int run = 0; run < (m); ++run) {
    cin >> u >> v >> c;
    from[run] = u;
    to[run] = v;
    cost[run] = c;
    V[make_pair(u, c)].push_back(v);
    V[make_pair(v, c)].push_back(u);
  }
  for (int run = 0; run < (m); ++run) {
    if (!have[make_pair(from[run], cost[run])]) go(from[run], cost[run]);
    if (!have[make_pair(to[run], cost[run])]) go(to[run], cost[run]);
  }
  int q;
  cin >> q;
  for (int run = 0; run < (q); ++run) {
    cin >> u >> v;
    if (u > v) swap(u, v);
    int res = f[make_pair(u, v)];
    for (int i = (1); i <= (coun); ++i) {
      res += (d[u][i] & d[v][i]);
    }
    cout << res << "\n";
  }
  return 0;
}
