/*

离线算法，LCATarjan

复杂度O(n+Q);

*/

#include<iostream>

#include<stdio.h>

#include<string.h>

using namespace std;

const int MAXN=;

const int MAXQ=;//查询数的最大值

int dis[MAXN];//到根节点的距离

//并查集部分

int F[MAXN];//需要初始化为-1

int find(int x){

    if(F[x]==-)return x;

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

}

void bing(int u,int v){

    int t1=find(u);

    int t2=find(v);

    if(t1!=t2)

        F[t1]=t2;

}

//***********************

bool vis[MAXN];//访问标记

int ancestor[MAXN];//祖先

struct Edge{

    int to,next;

    int d;

}edge[MAXN*];

int head[MAXN],tot;

void addedge(int u,int v,int d){

    edge[tot].to=v;

    edge[tot].d=d;

    edge[tot].next=head[u];

    head[u]=tot++;

}

struct Query{

    int q,next;

    int index;//查询编号

}query[MAXQ*];

int answer[MAXQ];//存储最后的查询结果，下标0 Q-1

int h[MAXN];//注意此处为MAXN...

int tt;

void add_query(int u,int v,int index){

    query[tt].q=v;

    query[tt].next=h[u];

    query[tt].index=index;

    h[u]=tt++;

    query[tt].q=u;

    query[tt].next=h[v];

    query[tt].index=index;

    h[v]=tt++;

}

void init(){

    tot=;

    memset(head,-,sizeof(head));

    tt=;

    memset(h,-,sizeof(h));

    memset(vis,false,sizeof(vis));

    memset(F,-,sizeof(F));

    memset(ancestor,,sizeof(ancestor));

}

void LCA(int u){

    ancestor[u]=u;

    vis[u]=true;

    for(int i=head[u];i!=-;i=edge[i].next){

        int v=edge[i].to;

        if(vis[v])continue;

        dis[v]=dis[u]+edge[i].d;

        LCA(v);

        bing(u,v);

        ancestor[find(u)]=u;

    }

    for(int i=h[u];i!=-;i=query[i].next){

        int v=query[i].q;

        if(vis[v]){

            answer[query[i].index]=ancestor[find(v)];

        }

    }

}

int main(){

    int n,m;

    int i;

    int u,v,d;

    char str[];

    init();

    scanf("%d%d",&n,&n);

    for(i=;i<n;++i){

        scanf("%d%d%d%1s",&u,&v,&d,str);

        addedge(u,v,d);

        addedge(v,u,d);

    }

    scanf("%d",&m);

    for(i=;i<m;++i){

        scanf("%d%d",&u,&v);

        add_query(u,v,i);

    }

    dis[]=;

    LCA();

    for(i=;i<m;++i){

        printf("%d\n",dis[query[i*].q]+dis[query[i*+].q]-*dis[answer[i]]);

    }

    return ;

}