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>
template <class T>
inline T qmin(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T qmax(T a, T b) {
  return a > b ? a : b;
}
const double eps = 1e-9;
const double PI = acos(-1);
const int inf = 0x3f3f3f3f;
const int mod = (int)1e9 + 7;
const int maxn = (int)1e5 + 5;
const long long INF = 0x3f3f3f3f3f3f3f3f;
using namespace std;
int n, m;
unordered_map<int, int> fa[maxn], out[maxn];
int sz[maxn];
int Find(int x, int c) {
  return fa[x][c] == x ? x : fa[x][c] = Find(fa[x][c], c);
}
void uni(int x, int y, int c) {
  if (fa[x].find(c) == fa[x].end()) fa[x][c] = x;
  if (fa[y].find(c) == fa[y].end()) fa[y][c] = y;
  int fx = Find(x, c), fy = Find(y, c);
  if (fx != fy) {
    if (sz[fx] < sz[fy])
      fa[fx][c] = fy, sz[fy] += sz[fx];
    else
      fa[fy][c] = fx, sz[fx] += sz[fy];
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (__typeof(n) i = 1; i <= (n); i++) sz[i] = 1;
  for (__typeof(m) i = 1; i <= (m); i++) {
    int u, v, c;
    scanf("%d %d %d", &u, &v, &c);
    uni(u, v, c);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    if (out[u].find(v) != out[u].end()) {
      printf("%d\n", out[u][v]);
      continue;
    }
    if (fa[u].size() > fa[v].size()) swap(u, v);
    int ans = 0;
    for (auto to : fa[u]) {
      int c = to.first;
      if (fa[v].find(c) != fa[v].end() && Find(u, c) == Find(v, c)) ans++;
    }
    out[u][v] = out[v][u] = ans;
    printf("%d\n", ans);
  }
  return 0;
}
