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 100002
using namespace std;
int tr[M*36],Lson[M*36],Rson[M*36],sum[M*36],cnt[M*36],tot,f[M][19],dep[M],n;
struct ll{
	int to,col,dis;
};
vector<ll>son[M];
void Insert(int &rt,int old,int l,int r,int col,int dis){
	rt=++tot;
	Lson[rt]=Lson[old];
	Rson[rt]=Rson[old];
	cnt[rt]=cnt[old]+1;
	sum[rt]=sum[old]+dis;
	if(l==r)return;
	int mid=(l+r)>>1;
	if(mid>=col)Insert(Lson[rt],Lson[old],l,mid,col,dis);
	else Insert(Rson[rt],Rson[old],mid+1,r,col,dis);
}
void dfs(int x,int fa,int col,int dis){
	dep[x]=dep[fa]+1,f[x][0]=fa;
	if(fa)Insert(tr[x],tr[fa],1,n-1,col,dis);
	for(int i=0;i<son[x].size();i++){
		ll y=son[x][i];
		if(y.to==fa)continue;
		dfs(y.to,x,y.col,y.dis);
	}
}
int Up(int b,int step){
	for(int i=0;i<17;i++)if(step&(1<<i))b=f[b][i];
	return b;
}
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=16;i>=0;i--)if(f[a][i]!=f[b][i])a=f[a][i],b=f[b][i];
	return f[a][0];
}
int query(int Rt,int rt,int Lt,int l,int r,int col,int dis){
	if(l==r){
		int num=cnt[Rt]+cnt[rt]-2*cnt[Lt];
		return num*dis;
	}
	int mid=(l+r)>>1;
	if(mid>=col)return query(Lson[Rt],Lson[rt],Lson[Lt],l,mid,col,dis)+sum[Rson[Rt]]+sum[Rson[rt]]-2*sum[Rson[Lt]];
	return sum[Lson[Rt]]+sum[Lson[rt]]-2*sum[Lson[Lt]]+query(Rson[Rt],Rson[rt],Rson[Lt],mid+1,r,col,dis);
}
int main(){
	int m,u,v,x,y;
	scanf("%d%d",&n,&m);
	for(int i=1;i<n;i++){
		scanf("%d%d%d%d",&u,&v,&x,&y);
		son[u].push_back({v,x,y});
		son[v].push_back({u,x,y});
	}
	dfs(1,0,0,0);
	for(int i=1;i<18;i++){
		if((1<<i)>n)break;
		for(int j=1;j<=n;j++)f[j][i]=f[f[j][i-1]][i-1];
	}
	for(int i=1;i<=m;i++){
		scanf("%d%d%d%d",&x,&y,&u,&v);
		int lca=LCA(u,v);
		printf("%d\n",query(tr[u],tr[v],tr[lca],1,n-1,x,y));
	}
	return 0;
}