Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define fastio                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define arrinit \
    int n;      \
    cin >> n;   \
    int arr[n]; \
    loop(i, n) { cin >> arr[i]; }
#define init(arr, n, val)       \
    for (int i = 0; i < n; i++) \
        arr[i] = val;
#define ret return
#define cont continue
#define endl "\n"
#define ll long long

#define loops(i, s, n) for (int i = s; i < n; i++)
#define loop(i, n) for (int i = 0; i < n; i++)
#define loopr(i, s, e) for (int i = s; i >= e; i--)
#define loopitr(mp) for (auto itr = mp.begin(); itr != mp.end(); itr++)

#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define pub push_back
#define pob pop_back
#define vpii vector<pair<int, int>>
#define vall(v) v.begin(), v.end()
#define mkp make_pair
#define ff first
#define ss second

#define cnt_bits __builtin_popcount
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define eps 1e-9
const int mod = 1000000007;
set<int> tot;



int main()
{
    fastio;
    int n , m;
    cin>>n>>m;
    map<int,set<int>> mp;
    loop(i, m){
        int u, v, c;
        cin>>u>>v>>c;
        u--, v--;
        mp[c].insert(u);
        mp[c].insert(v);
    }

    int q;
    cin>>q;
    loop(i, q){
        int u, v;
        cin>>u>>v; u--, v--;
        int cnt = 0;
        for(auto itr : mp) {
            auto st = itr.ss;
            if(st.find(u) != st.end() && st.find(v) != st.end()){
                cnt++;
            }
        }
        cout<<cnt<<endl;
    }
}
