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 N 100005
using namespace std;
int n,m,dep[N],fa[N][25];
struct data{
	int to,col,v;
};
vector<data>a[N];
int disx[N],disy[N],disz[N],X[N],Y[N],Z[N],Col[N],V[N],C[N];
int cnt[N],sum[N];
vector<int>Q[N];
void dfs(int x,int f){
	dep[x]=dep[f]+1;
	fa[x][0]=f;
	for(int i=0;i<(int)a[x].size();i++){
		int y=a[x][i].to;
		if(y==f)continue;
		dfs(y,x);
	}
}
int Up(int x,int step){
	for(int i=0;i<=20;i++)
		if(step&(1<<i))x=fa[x][i];
	return x;
}
int LCA(int x,int y){
	if(dep[x]>dep[y])swap(x,y);
	y=Up(y,dep[y]-dep[x]);
	if(x==y)return x;
	for(int i=20;i>=0;i--)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];
	return fa[x][0];
}
void dfs1(int x,int d){
	for(int i=0;i<(int)Q[x].size();i++){
		int y=Q[x][i];
		if(x==X[y])disx[y]=d-sum[Col[y]]+cnt[Col[y]]*C[y];
		if(x==Y[y])disy[y]=d-sum[Col[y]]+cnt[Col[y]]*C[y];
		if(x==Z[y])disz[y]=d-sum[Col[y]]+cnt[Col[y]]*C[y];
	}
	for(int i=0;i<(int)a[x].size();i++){
		int y=a[x][i].to,col=a[x][i].col,v=a[x][i].v;
		if(y==fa[x][0])continue;
		cnt[col]++;sum[col]+=v;
		dfs1(y,d+v);
		cnt[col]--;sum[col]-=v;
	}
}
signed main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<n;i++){
		int x,y,col,v;
		scanf("%d%d%d%d",&x,&y,&col,&v);
		a[x].push_back((data){y,col,v});
		a[y].push_back((data){x,col,v});
	}
	dfs(1,0);
	for(int i=1;(1<<i)<=n;i++)
		for(int j=1;j<=n;j++)fa[j][i]=fa[fa[j][i-1]][i-1];
	for(int i=1;i<=m;i++){
		scanf("%d%d%d%d",&Col[i],&C[i],&X[i],&Y[i]);
		Q[X[i]].push_back(i);
		Q[Y[i]].push_back(i);
		Z[i]=LCA(X[i],Y[i]);
		Q[Z[i]].push_back(i);
	}
	dfs1(1,0);
	for(int i=1;i<=m;i++)printf("%d\n",disx[i]+disy[i]-2*disz[i]);
	return 0;
}