// clang-format off
#include<bits/stdc++.h>
#define pb push_back
#define P make_pair
#define st first
#define nd second
#define bit(s,x) (((s)>>(x))&1)
#define ctz(s) __builtin_popcount(s)
using namespace std;
typedef long long ll; 
typedef unsigned long long ull; 
//typedef __int128 i128; 
typedef long double ld;
typedef pair<int,int> pii; typedef pair<ll,int> pli; typedef pair<ll,ll> pll; typedef pair<int,ll> pil;
inline ll read(){
	ll x=0,f=1,c=getchar();
	while(c<'0'||c>'9')f=(c=='-'?-1:1),c=getchar();
	while(c>='0'&&c<='9')x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
inline void write(ll x){ if(x<0) putchar('-'),x=-x; if(x>9) write(x/10); putchar(x%10+'0'); }
// remember to change!!!
const int mod=998244353;
inline void plust(int &x,int y,int p=mod){x=x+y<p?x+y:x+y-p;}
inline void mul(int &x,int y,int p=mod){x=1ll*x*y%p;}
inline int fp(int x,ll p=mod-2,int m=mod,int res=1){
	for(;p;p>>=1){if(p&1)mul(res,x,m);mul(x,x,m);}
	return res;
}
const int N=1e5+10;
const int INF=0x3fffffff;
// clang-format on
int n,m;
vector<pil>G1[N],G2[N];
vector<pii>q[N];
int dfn1[N],dfn2[N],idfn[N],st1[17][N],siz[N],st[17][N],tim;
ll dep1[N],dep2[N],ans[N*5];
inline int pmin(int a,int b){return dfn2[a] < dfn2[b]?a:b;}
inline int pmin1(int a,int b){return dfn1[a] < dfn1[b]?a:b;}
inline void dfs1(int u,int fa){
	st[0][dfn2[u]=++tim]=fa;
	for(auto [v,w]:G2[u])if(v^fa){
		dep2[v] = dep2[u] + w;
		dfs1(v,u);
	}
}
inline ll dist2(pil A,pil B){
	int u=A.st,v=B.st;
	if(u==v)return A.nd;
	ll res = dep2[u] + dep2[v];
	if((u=dfn2[u]) > (v=dfn2[v]))swap(u,v);
	int k=__lg(v-u++);
	int lca = pmin(st[k][u],st[k][v-(1<<k)+1]);
	return res - 2*dep2[lca] + A.nd + B.nd;
}
inline ll dist1(int u,int v){
	if(u==v)return 0;
	ll res = dep1[u] +dep1[v];
	if((u=dfn1[u]) > (v=dfn1[v]))swap(u,v);
	int k=__lg(v-u++);
	int lca = pmin1(st1[k][u],st1[k][v-(1<<k)+1]);
	return res - 2*dep1[lca];
}
inline void dfs2(int u,int fa){
	st1[0][dfn1[u] = ++tim]=fa;idfn[tim]=u;
	siz[u]=1;
	for(auto [v,w]:G1[u])if(v^fa){
		dep1[v] = dep1[u] + w;
		dfs2(v,u);
		siz[u] += siz[v];
	}
}
typedef pair<pil,pil> info;
info t[1<<18];
ll tag[1<<18];
inline info merge(info A,info B){
	pil a[4] = {A.st,A.nd,B.st,B.nd};
	info res = A;
	for(int i=0;i<4;++i)for(int j=i+1;j<4;++j)if(dist2(a[i],a[j]) > dist2(res.st,res.nd))res = P(a[i],a[j]);
	return res;
}
#define mid ((l+r)>>1)
inline void update(int x,int l,int r,int nl,int nr,ll k){
	if(nl>nr)return;
	if(nl<=l && r<=nr){
		tag[x]+=k;
		t[x].st.nd+=k,t[x].nd.nd+=k;
		return ;
	}
	tag[x<<1]+=tag[x];tag[x<<1|1]+=tag[x];
	t[x<<1].st.nd+=tag[x];
	t[x<<1].nd.nd+=tag[x];
	t[x<<1|1].st.nd+=tag[x];
	t[x<<1|1].nd.nd+=tag[x];
	tag[x]=0;
	if(nl<=mid)update(x<<1,l,mid,nl,nr,k);
	if(mid<nr) update(x<<1|1,mid+1,r,nl,nr,k);
	t[x] = merge(t[x<<1],t[x<<1|1]);
}
inline void build(int x,int l,int r){
	if(l==r){
		int u=idfn[l];
		t[x] = P(P(u,dep1[u]),P(u,dep1[u]));
		return;
	}
	build(x<<1,l,mid),build(x<<1|1,mid+1,r);
	t[x] = merge(t[x<<1],t[x<<1|1]);
}
inline void dfs3(int u,int fa){
	int d1=t[1].st.st,d2=t[1].nd.st;
	for(auto [v,id]:q[u]) ans[id] = max(dist1(u,d1)+dist2(P(v,0),P(d1,0)),dist1(u,d2)+dist2(P(v,0),P(d2,0)));
	for(auto [v,w]:G1[u])if(v^fa){
		update(1,1,n,dfn1[v],dfn1[v]+siz[v]-1,-w);
		update(1,1,n,1,dfn1[v]-1,w);
		update(1,1,n,dfn1[v]+siz[v],n,w);
		dfs3(v,u);
		update(1,1,n,dfn1[v],dfn1[v]+siz[v]-1,w);
		update(1,1,n,1,dfn1[v]-1,-w);
		update(1,1,n,dfn1[v]+siz[v],n,-w);
	}
}
int main() {
    //freopen("data.in", "r", stdin);
    // freopen(".in","r",stdin);
    // freopen("myans.out","w",stdout);
	n=read(),m=read();
	for(int i=1;i<n;++i){
		int u=read(),v=read(),w=read();
		G1[u].pb(P(v,w));
		G1[v].pb(P(u,w));
	}
	for(int i=1;i<n;++i){
		int u=read(),v=read(),w=read();
		G2[u].pb(P(v,w));
		G2[v].pb(P(u,w));
	}
	for(int i=1;i<=m;++i){
		int a=read(),b=read();
		q[a].pb(P(b,i));
	}
	tim=0,dfs1(1,0);
	for(int i=1;i<17;++i)for(int j=1;j+(1<<i)-1<=n;++j)st[i][j]=pmin(st[i-1][j],st[i-1][j+(1<<(i-1))]);
	tim=0,dfs2(1,0);
	build(1,1,n);
	for(int i=1;i<17;++i)for(int j=1;j+(1<<i)-1<=n;++j)st1[i][j]=pmin1(st1[i-1][j],st1[i-1][j+(1<<(i-1))]);
	dfs3(1,0);
	for(int i=1;i<=m;++i)printf("%d\n",ans[i]);
    return 0;
}