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 M 100005
using namespace std; 
int fa[M][20],dep[M],dis[M];
struct node{
	int to,c,v;
};
vector<node>edge[M];
struct Query{
	int id,c,d,flag;
};
int cc[M],cs[M];
vector<Query>query[M];
int ans[M];
void dfs(int x,int f,int d,int ds){
	fa[x][0]=f;dep[x]=d;dis[x]=ds;
	for(int i=0;i<edge[x].size();i++){
		int y=edge[x][i].to;
		if(y==f)continue;
		dfs(y,x,d+1,ds+edge[x][i].v);
	}
}
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 a,int b){
	if(dep[a]>dep[b])swap(a,b);
	b=Up(b,dep[b]-dep[a]);
	if(a==b)return a;
	for(int i=19;i>=0;i--)
		if(fa[a][i]!=fa[b][i])
			a=fa[a][i],b=fa[b][i];
	return fa[a][0];
}
void DFS(int x,int f){
	for(int i=0;i<query[x].size();i++){
		int c=query[x][i].c;
		int d=query[x][i].d;
		ans[query[x][i].id]+=query[x][i].flag*(dis[x]-cs[c]+cc[c]*d);
	}
	for(int i=0;i<edge[x].size();i++){
		int y=edge[x][i].to;
		if(y==f)continue;
		cc[edge[x][i].c]++;
		cs[edge[x][i].c]+=edge[x][i].v;
		DFS(y,x);
		cc[edge[x][i].c]--;
		cs[edge[x][i].c]-=edge[x][i].v; 
	}
}
int main(){
	int n,Q,a,b,c,d;
	cin>>n>>Q;
	for(int i=1;i<n;i++){
		scanf("%d %d %d %d",&a,&b,&c,&d);
		edge[a].push_back((node){b,c,d});
		edge[b].push_back((node){a,c,d}); 
	}
	dfs(1,0,0,0);
	for(int j=1;j<20;j++)
		for(int i=1;i<=n;i++)fa[i][j]=fa[fa[i][j-1]][j-1];
	for(int i=1;i<=Q;i++){
		scanf("%d %d %d %d",&c,&d,&a,&b);
		query[a].push_back((Query){i,c,d,1});
		query[b].push_back((Query){i,c,d,1});
		query[LCA(a,b)].push_back((Query){i,c,d,-2}); 
//		printf("a=%d b=%d c=%d\n",a,b,LCA(a,b));
	}
	DFS(1,0); 
	for(int i=1;i<=Q;i++)printf("%d\n",ans[i]);
	return 0;
}