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;
using ll = long long;
using db = double;
using ld = long double;
const ll mod = 1000000007;
const ll INF = 1000000099;
const ll LINF = (ll)(1e18 + 99);
vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
template <typename T, typename U>
void chmin(T& t, const U& u) {
  if (t > u) t = u;
}
template <typename T, typename U>
void chmax(T& t, const U& u) {
  if (t < u) t = u;
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T mpow(T a, T n) {
  T res = 1;
  for (; n; n >>= 1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}
template <typename T>
class UnionFind {
 public:
  vector<T> Parent;
  UnionFind(T N) { Parent = vector<T>(N, -1); }
  T root(T A) {
    if (Parent[A] < 0) return A;
    return Parent[A] = root(Parent[A]);
  }
  T size(T A) { return -Parent[root(A)]; }
  bool connect(T A, T B) {
    A = root(A);
    B = root(B);
    if (A == B) {
      return false;
    }
    if (size(A) < size(B)) swap(A, B);
    Parent[A] += Parent[B];
    Parent[B] = A;
    return true;
  }
  bool same(T A, T B) { return root(A) == root(B); }
};
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<UnionFind<ll>> v(m, UnionFind<ll>(n));
  for (ll i = 0; i < (ll)(m); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    c--;
    v[c].connect(a, b);
  }
  int q;
  cin >> q;
  while (q--) {
    int x, y, ans = 0;
    cin >> x >> y;
    x--;
    y--;
    for (ll i = 0; i < (ll)(m); i++) {
      if (v[i].same(x, y)) ans++;
    }
    cout << ans << endl;
  }
}
