Problem: There is a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i, and the color and length of that edge are c_i and d_i, respectively. Here the color of each edge is represented by an integer between 1 and N-1 (inclusive). The same integer corresponds to the same color, and different integers correspond to different colors.

Answer the following Q queries:

* Query j (1 \leq j \leq Q): assuming that the length of every edge whose color is x_j is changed to y_j, find the distance between Vertex u_j and Vertex v_j. (The changes of the lengths of edges do not affect the subsequent queries.)

Constraints

* 2 \leq N \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq a_i, b_i \leq N
* 1 \leq c_i \leq N-1
* 1 \leq d_i \leq 10^4
* 1 \leq x_j \leq N-1
* 1 \leq y_j \leq 10^4
* 1 \leq u_j < v_j \leq N
* The given graph is a tree.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
a_1 b_1 c_1 d_1
:
a_{N-1} b_{N-1} c_{N-1} d_{N-1}
x_1 y_1 u_1 v_1
:
x_Q y_Q u_Q v_Q


Output

Print Q lines. The j-th line (1 \leq j \leq Q) should contain the answer to Query j.

Example

Input

5 3
1 2 1 10
1 3 2 20
2 4 4 30
5 2 1 40
1 100 1 4
1 100 1 5
3 1000 3 4


Output

130
200
60

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define For(a,b,c) for(int a=b;a<=c;++a)
#define Dor(a,b,c) for(int a=b;a>=c;--a)
using namespace std;
const int N=100007;
int n,m,hd[N],nt[N<<1],to[N<<1],C[N<<1],D[N<<1],A[N],B[N],Qx[N],Qy[N],Qz[N],Qc[N],Qd[N],Dx[N],Dy[N],Dz[N],fa[N],son[N],sz[N],tp[N],de[N];
vector<int> V[N];
void walk(int u) {
	sz[u]=1,de[u]=de[fa[u]]+1;
	for(int i=hd[u];i;i=nt[i])
	if(to[i]!=fa[u]) {
		fa[to[i]]=u;
		walk(to[i]);
		sz[u]+=sz[to[i]];
		if(sz[to[i]]>sz[son[u]]) son[u]=to[i];
	}
}
void run(int u) {
	if(son[u]) tp[son[u]]=tp[u],run(son[u]);
	for(int i=hd[u];i;i=nt[i])
	if(to[i]!=fa[u]&&to[i]!=son[u]) tp[to[i]]=to[i],run(to[i]);
}
int lca(int x,int y) {
	while(tp[x]!=tp[y]) {
		if(de[tp[x]]>de[tp[y]]) x=fa[tp[x]];
		else y=fa[tp[y]];
	}
	return de[x]<de[y]?x:y;
}
void fly(int u,int d) {
	For(i,0,(int)V[u].size()-1) {
		int q=V[u][i];
		if(u==Qx[q]) Dx[q]=d-B[Qc[q]]+A[Qc[q]]*Qd[q];
		if(u==Qy[q]) Dy[q]=d-B[Qc[q]]+A[Qc[q]]*Qd[q];
		if(u==Qz[q]) Dz[q]=d-B[Qc[q]]+A[Qc[q]]*Qd[q];
	}
	for(int i=hd[u];i;i=nt[i])
	if(to[i]!=fa[u]) {
		++A[C[i]],B[C[i]]+=D[i];
		fly(to[i],d+D[i]);
		--A[C[i]],B[C[i]]-=D[i];
	}
}
int main() {
	scanf("%d%d",&n,&m);
	For(i,1,n-1) {
		int x,y,c,d;
		scanf("%d%d%d%d",&x,&y,&c,&d);
		nt[i*2-1]=hd[x],hd[x]=i*2-1,to[i*2-1]=y,C[i*2-1]=c,D[i*2-1]=d;
		nt[i*2]=hd[y],hd[y]=i*2,to[i*2]=x,C[i*2]=c,D[i*2]=d;
	}
	walk(1),tp[1]=1,run(1);
	For(i,1,m) {
		scanf("%d%d%d%d",&Qc[i],&Qd[i],&Qx[i],&Qy[i]);
		Qz[i]=lca(Qx[i],Qy[i]);
		V[Qx[i]].push_back(i);
		V[Qy[i]].push_back(i);
		V[Qz[i]].push_back(i);
	}
	fly(1,0);
	For(i,1,m) printf("%d\n",Dx[i]+Dy[i]-2*Dz[i]);
	return 0;
}