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>
double const EPS = 3e-8;
using namespace std;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
inline bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
const int MX = 1005;
vector<pair<int, int> > adj[MX];
int n, m, q;
int Ans[MX][MX];
int parnt[105][105];
int ok[105][105][105];
int vis[105][105], cs;
int dfs(int cur, int lok, int c) {
  if (cur == lok) return 1;
  vis[cur][c] = cs;
  int i;
  for (i = 1; i <= n; i++) {
    if (i == cur || vis[i][c] == cs || ok[cur][i][c] == 0) continue;
    if (ok[cur][i][c] == 1)
      if (dfs(i, lok, c)) return 1;
  }
  return 0;
}
int main() {
  n = ({
    int a;
    read(a);
    a;
  }),
  m = ({
    int a;
    read(a);
    a;
  });
  for (__typeof((m)-1) i = (0); i <= (m)-1; ++i) {
    int x = ({
      int a;
      read(a);
      a;
    });
    int y = ({
      int a;
      read(a);
      a;
    });
    int c = ({
      int a;
      read(a);
      a;
    });
    ok[x][y][c] = ok[y][x][c] = 1;
  }
  q = ({
    int a;
    read(a);
    a;
  });
  while (q--) {
    cs = q + 1;
    int x = ({
      int a;
      read(a);
      a;
    });
    int y = ({
      int a;
      read(a);
      a;
    });
    int ans = 0;
    for (__typeof(m) i = (1); i <= m; ++i) {
      int add = dfs(x, y, i);
      if (add) {
        ans++;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
