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 - a_i, b_i(1 ≤ a_i < b_i ≤ n) and c_i(1 ≤ c_i ≤ 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, (a_i, b_i, c_i) ≠ (a_j, b_j, c_j).

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 - u_i and v_i(1 ≤ u_i, v_i ≤ n). It is guaranteed that u_i ≠ v_i.

Output

For each query, print the answer in a separate line.

Examples

4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4

2
1
0

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

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<bits/stdc++.h>

#define pii pair<int,int>

#define mp make_pair

#define F first

#define S second

#define rep(i,a,b) for(int i=a;i<=b;i++)

using namespace std;

const int maxn=;

int Laxt[maxn],Next[maxn],To[maxn],id[maxn],ans[maxn],cnt;

int a[maxn],b[maxn],fa[maxn],times[maxn],T,q[maxn],tot,N;

map<pii,int>Mp;

map<pii,int>fcy;

struct in{

    int u,v,col;

    bool friend operator <(in w,in v){ return w.col<v.col; }

}s[maxn];

void add(int u,int v,int o)

{

    Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v; id[cnt]=o;

}

int find(int x){

    if(x==fa[x]) return x;

    return fa[x]=find(fa[x]);

}

void merge(int x,int y)

{

    int fx=find(x),fy=find(y);

    fa[fx]=fy;

}

int main()

{

    int M,Q,u,v;

    scanf("%d%d",&N,&M);

    rep(i,,M) scanf("%d%d%d",&s[i].u,&s[i].v,&s[i].col);

    sort(s+,s+M+);

    scanf("%d",&Q);

    rep(i,,Q){

        scanf("%d%d",&a[i],&b[i]); if(a[i]>b[i]) swap(a[i],b[i]);

        add(a[i],b[i],i);

        fcy[mp(a[i],b[i])]=;

    }

    rep(i,,M){

        int j=i; T++; merge(s[i].u,s[i].v);

        while(j+<=M&&s[j+].col==s[i].col) j++;

        tot=;

        rep(k,i,j) q[++tot]=s[k].u,q[++tot]=s[k].v;

        sort(q+,q+tot+); tot=unique(q+,q+tot+)-(q+);

        rep(k,,tot) fa[q[k]]=q[k],times[q[k]]=T;

        rep(k,i,j) merge(s[k].u,s[k].v);

        if(tot>sqrt(N)) rep(k,,tot) {

            for(int w=Laxt[q[k]];w;w=Next[w]){

                if(times[To[w]]==T&&find(q[k])==find(To[w])) ans[id[w]]++;

            }

        }

        else {

            rep(k,,tot)

             rep(p,k+,tot){

                  if(find(q[k])==find(q[p])&&fcy.find(mp(q[k],q[p]))!=fcy.end()) Mp[mp(q[k],q[p])]++;

            }

        }

        i=j;

    }

    rep(i,,Q) printf("%d\n",ans[i]+Mp[mp(a[i],b[i])]);

    return ;

}