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;
int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
int kx[] = {2, 2, -2, -2, 1, 1, -1, -1};
int ky[] = {1, -1, 1, -1, 2, -2, 2, -2};
void dance() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void file() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
template <typename t>
void printV(vector<t> v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << ' ';
  }
  cout << endl;
}
bool isPrime(long long n) {
  if (n == 2) return true;
  if (n < 2 || n % 2 == 0) return false;
  for (long long i = 3; i * i <= n; i += 2)
    if (n % i == 0) return false;
  return true;
}
bool prime[100001];
void seive() {
  memset(prime, true, sizeof(prime));
  prime[1] = false;
  for (int i = 2; i <= 100000; i++) {
    if (prime[i])
      for (int j = i + i; j <= 100000; j += i) {
        prime[j] = false;
      }
  }
}
long long MOD = 1000000007;
int p[101][101];
int parent(int c, int x) {
  if (p[c][x] == x) return x;
  return p[c][x] = parent(c, p[c][x]);
}
bool sameset(int c, int x, int y) { return (parent(c, x) == parent(c, y)); }
void mer(int c, int x, int y) { p[c][parent(c, x)] = parent(c, y); }
int main() {
  dance();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++) p[i][j] = j;
  }
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    mer(c, a, b);
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, r = 0;
    cin >> a >> b;
    for (int i = 1; i <= m; i++) r += sameset(i, a, b);
    cout << r << endl;
  }
}
