#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=40000+100;
int n,m,k1,k2,t;
int hade1[maxn],hade2[maxn];
int d[maxn],c[maxn];
bool vis[maxn];
struct note
{    int fr;int next;int to;int len;int lca;
}aa[maxn*2],bb[1000];
void addage1(int fr,int to,int len)
{aa[k1].to=to;aa[k1].len=len;aa[k1].next=hade1[fr];hade1[fr]=k1++;
}
void addage2(int fr,int to)
{bb[k2].fr=fr;bb[k2].to=to;bb[k2].lca=-1;bb[k2].next=hade2[fr];hade2[fr]=k2++;
}
void init()
{memset(hade1,-1,sizeof(hade1));memset(hade2,-1,sizeof(hade2));memset(d,0,sizeof(d));memset(vis,0,sizeof(vis));k1=k2=0;
}
int found(int x)
{if(x==c[x]) return x;else return c[x]=found(c[x]);
}
void unit(int x,int y)
{x=found(x);y=found(y);if(x!=y)c[y]=x;
}
void tarjan(int u)
{vis[u]=1;c[u]=u;for(int i=hade1[u];i!=-1;i=aa[i].next){int v=aa[i].to;int w=aa[i].len;if(!vis[v]){d[v]=d[u]+w;//从此节点到根节点的距离
             tarjan(v);unit(u,v);}}for(int i=hade2[u];i!=-1;i=bb[i].next){int v=bb[i].to;if(vis[v]){bb[i].lca=bb[i^1].lca=found(v);//回溯的是u和v最近公共子节点的分支，所以v是指向最近公共子节点的
          }}}
int main()
{scanf("%d",&t);while(t--){   init();scanf("%d%d",&n,&m);for(int i=1;i<n;i++){int fr,to,len;scanf("%d%d%d",&fr,&to,&len);addage1(fr,to,len);addage1(to,fr,len);}for(int i=0;i<m;i++){int fr,to;scanf("%d%d",&fr,&to);addage2(fr,to);addage2(to,fr);}d[1]=0;tarjan(1);for(int i=0;i<m;i++){int s=i*2;int u=bb[s].fr,v=bb[s].to,lca=bb[s].lca;int mm=d[u]+d[v]-2*d[lca];printf("%d\n",mm);}}return 0;
}