Mr. Kitayuta has just bought an undirected graph consisting of n vertices and medges. 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 — aibi (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.

 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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath> const int maxn=1e5+;
typedef long long ll;
using namespace std;
struct node
{
int v;
int val;
};
int ans=;
int vis[];
vector<node>vec[];
void bfs(int x,int y)
{
node start;
for(int j=;j<=;j++)
{
queue<node>q;
memset(vis,,sizeof(vis));
vis[x]=;
for(int t=;t<vec[x].size();t++)
{
if(vec[x][t].val==j&&vis[vec[x][t].v]==)
{
q.push(vec[x][t]);
vis[vec[x][t].v]=;
}
}
while(!q.empty())
{
node now=q.front();
q.pop();
if(now.v==y)
{
ans++;
}
for(int t=;t<vec[now.v].size();t++)
{
if(vec[now.v][t].val==j&&vis[vec[now.v][t].v]==)
{
vis[vec[now.v][t].v]=;
q.push(vec[now.v][t]);
}
}
}
}
}
int main()
{
int n,m;
cin>>n>>m; int a,b,c;
set<int>s;
for(int t=;t<m;t++)
{
scanf("%d%d%d",&a,&b,&c);
s.insert(c);
node st;
st.v=b;
st.val=c;
vec[a].push_back(st);
st.v=a;
st.val=c;
vec[b].push_back(st);
}
int q;
cin>>q;
int uu,vv;
for(int t=;t<q;t++)
{
scanf("%d%d",&uu,&vv);
ans=;
bfs(uu,vv);
printf("%d\n",ans);
} return ;
}

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