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>
using namespace std;
int n,K;
#define M 100005
struct Edge{
	int to,nx,id,v;
}edge[M<<1];
int h[M],tt;
void link(int a,int b,int id,int v){
	edge[++tt].to=b;
	edge[tt].nx=h[a];
	edge[tt].id=id;
	edge[tt].v=v;
	h[a]=tt;
}
#define DEG(i,st) for(int i=h[st];i;i=edge[i].nx)
#define LL long long
int fa[M][19],dep[M];
LL Ans[M];
LL dis[M];
void DFS(int F,int now){
	dep[now]=dep[F]+1;
	fa[now][0]=F;
	for(int i=1;i<19;i++)
		fa[now][i]=fa[fa[now][i-1]][i-1];
	DEG(i,now){
		int to=edge[i].to;
		if(to==F)continue;
		dis[to]=dis[now]+edge[i].v;
		DFS(now,to);
	}	
}
struct Node{
	int id,col,f,to;
};
vector<Node>Q[M];
int LCA(int a,int b){
	if(dep[a]>dep[b])swap(a,b);
	int stp=dep[b]-dep[a];
	for(int i=0;i<19;i++)
		if((1<<i)&stp)b=fa[b][i];
	if(a==b)return a;
	for(int i=18;~i;--i)
		if(fa[a][i]!=fa[b][i])
			a=fa[a][i],b=fa[b][i];
	return fa[a][0];
}
int V[M],ID[M];
int cnt[M];
LL Cnt[M];
void dfs(int F,int now){
	if(F){
		Cnt[ID[now]]+=V[now];
		cnt[ID[now]]++;
	}
	DEG(i,now){
		int to=edge[i].to;
		if(to==F)continue;
		V[to]=edge[i].v;
		ID[to]=edge[i].id;
		dfs(now,to);
	}
	for(int i=0;i<Q[now].size();i++){
		Node Now=Q[now][i];
		LL d=dis[now];
		d-=Cnt[Now.col];
		d+=1LL*cnt[Now.col]*Now.to;
		Ans[Now.id]+=d*Now.f;
	}
	if(F){
		Cnt[ID[now]]-=V[now];
		cnt[ID[now]]--;
	}
}
int main(){
	cin>>n>>K;
	for(int i=1,a,b,c,d;i<n;i++){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		link(a,b,c,d),link(b,a,c,d);
	}
	DFS(0,1);
	for(int x,y,u,v,i=1;i<=K;i++){
		scanf("%d%d%d%d",&x,&y,&u,&v);
		int l=LCA(u,v);
		Q[u].push_back((Node){i,x,1,y});
		Q[v].push_back((Node){i,x,1,y});
		Q[l].push_back((Node){i,x,-2,y});
	}
	dfs(0,1);
	for(int i=1;i<=K;i++)
		printf("%lld\n",Ans[i]);
	return 0;
}