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 int long long
using namespace std;
const int MAXN=3e5+10;
int n,pre[MAXN][20],lg[MAXN],deep[MAXN],m,dis,sum[MAXN],ans[MAXN],tot[MAXN];
struct node{
	int ed,v,co;
};
vector<node> G[MAXN];
struct que{
	int pos,co,val,id;
};
vector<que> Q[MAXN];
void DFS0(int x,int fa){
	//cout<<dis<<" "<<x<<endl;
	for(int i=0;i<Q[x].size();i++){
		ans[Q[x][i].id]+=Q[x][i].pos*(dis-tot[Q[x][i].co]+sum[Q[x][i].co]*Q[x][i].val);
	}
	for(int i=0;i<G[x].size();i++){
		int t=G[x][i].ed;
		if(t==fa)continue;
		dis+=G[x][i].v;
		sum[G[x][i].co]++;
		tot[G[x][i].co]+=G[x][i].v;
		DFS0(t,x);
		dis-=G[x][i].v;
		sum[G[x][i].co]--;
		tot[G[x][i].co]-=G[x][i].v;
	}
}
void DFS(int x,int fa){
	pre[x][0]=fa;
	deep[x]=deep[fa]+1;
	for(int i=1;(1<<i)<=deep[x];i++)pre[x][i]=pre[pre[x][i-1]][i-1];
	for(int i=0;i<G[x].size();i++){
		int t=G[x][i].ed;
		if(t==fa)continue;
		DFS(t,x);
	}
}
int LCA(int x,int y){
	if(deep[x]<deep[y])swap(x,y);
	while(deep[x]>deep[y])x=pre[x][lg[deep[x]-deep[y]]-1];
	if(x==y)return x;
	//cout<<x<<" "<<y<<endl;
	for(int i=lg[deep[x]]-1;i>=0;i--){
		if(pre[x][i]==pre[y][i])continue;
		x=pre[x][i],y=pre[y][i];
	}
	return pre[x][0];
}
signed main(){
	scanf("%lld %lld",&n,&m);
	for(int i=1;i<=MAXN-10;i++)lg[i]=lg[i-1]+((1<<lg[i-1])==i);
	for(int i=1;i<=n-1;i++){
		int st,ed,v,co;
		scanf("%lld %lld %lld %lld",&st,&ed,&co,&v);
		G[st].push_back(node{ed,v,co});
		G[ed].push_back(node{st,v,co});
	}
	DFS(1,0);
	for(int i=1;i<=m;i++){
		int co,val,x,y;
		scanf("%lld %lld %lld %lld",&co,&val,&x,&y);
		Q[x].push_back(que{1,co,val,i});
		Q[y].push_back(que{1,co,val,i});
		Q[LCA(x,y)].push_back(que{-2,co,val,i});
		//cout<<LCA(x,y)<<endl;
	}
	DFS0(1,0);
	for(int i=1;i<=m;i++)printf("%lld\n",ans[i]);
	return 0;
}