Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int mxn = 233;
const int mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) f = c == '-' ? -1 : 1, c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * f;
}
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long ksm(long long a, long long b, long long mod) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long inv2(long long a, long long mod) { return ksm(a, mod - 2, mod); }
void exgcd(long long a, long long b, long long &x, long long &y, long long &d) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, y, x, d);
    y -= x * (a / b);
  }
}
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int n, m, f[mxn][mxn];
void init() {
  for (int i = 1; i < mxn; ++i)
    for (int j = 1; j < mxn; ++j) f[i][j] = j;
}
int find(int u, int c) {
  int rt = u;
  while (f[c][rt] != rt) rt = f[c][rt];
  int y = u;
  while (y != rt) {
    int t = f[c][y];
    f[c][y] = rt;
    y = t;
  }
}
void join(int u, int v, int c) {
  int fu = find(u, c), fv = find(v, c);
  if (fu != fv) f[c][fu] = fv;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  ;
  cin >> n >> m;
  init();
  int u, v, c;
  for (int i = 1; i <= m; ++i) {
    cin >> u >> v >> c;
    join(u, v, c);
  }
  int q;
  cin >> q;
  while (q--) {
    cin >> u >> v;
    int res = 0;
    for (int i = 1; i <= m; ++i)
      if (find(u, i) == find(v, i)) res++;
    cout << res << endl;
  }
  return 0;
}
