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>
inline T sqr(const T& x) {
  return x * x;
}
template <class T>
inline void updMin(T& a, const T& b) {
  if (b < a) a = b;
}
template <class T>
inline void updMax(T& a, const T& b) {
  if (b > a) a = b;
}
const int maxn = 200000;
int n, m;
vector<int> a[maxn], c[maxn];
map<pair<int, int>, int> used;
int usedcol[maxn];
int ITER;
unordered_map<int, int> p[maxn];
inline int getv(int v, int c) {
  if (p[v].count(c)) {
    int t = p[v][c];
    if (t != v)
      return p[v][c] = getv(t, c);
    else
      return t;
  } else {
    return v;
  }
}
void unite(int c, int a, int b) {
  a = getv(a, c);
  b = getv(b, c);
  if (rand() % 2) swap(a, b);
  p[a][c] = b;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    int x, y, z;
    scanf("%d %d %d", &x, &y, &z);
    x--, y--;
    a[x].push_back(y);
    a[y].push_back(x);
    c[x].push_back(z);
    c[y].push_back(z);
    unite(z, x, y);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--, y--;
    if (x > y) swap(x, y);
    if (used.count(make_pair(x, y))) {
      printf("%d\n", used[make_pair(x, y)]);
      continue;
    }
    int ans = 0;
    ITER++;
    int mem = x;
    if (c[y].size() < c[x].size()) mem = y;
    for (int j = 0; j < c[mem].size(); j++) {
      int col = c[mem][j];
      if (usedcol[col] == ITER) continue;
      usedcol[col] = ITER;
      if (getv(x, col) == getv(y, col)) ans++;
    }
    used[make_pair(x, y)] = ans;
    printf("%d\n", ans);
  }
  return 0;
}
