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;
const int N=1e5+5;
int n,m;
int fa[N][20],dep[N];
int dis[N],cnt[N],ret[N];
struct node{int u,co,l;};
vector<node>e[N];
struct ques{int id,co,d,k;};
vector<ques>q[N];
inline int read(){
    int x=0;char c=getchar();
    while(c<'0'||c>'9')c=getchar();
    while(c>='0'&&c<='9')x=(x<<3)+(x<<1)+(c^48),c=getchar();
    return x;
}
void pre(int x,int f){
	fa[x][0]=f,dep[x]=dep[f]+1;
	for(int i=0;i<e[x].size();i++)
		if(e[x][i].u!=f)pre(e[x][i].u,x);
}
int UP(int x,int step){
    for(int i=0;i<=17;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=17;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,int D){
	for(int i=0;i<q[x].size();i++){
		ques now=q[x][i];
		ret[now.id]+=now.k*(D-dis[now.co]+cnt[now.co]*now.d);
	}
	for(int i=0;i<e[x].size();i++){
		node y=e[x][i];
		if(y.u==f)continue;
		cnt[y.co]++,dis[y.co]+=y.l;
		dfs(y.u,x,D+(y.l));
		cnt[y.co]--,dis[y.co]-=y.l;
	}
}
int main(){
    n=read(),m=read();
    for(int i=1,a,b,c,d;i<n;i++){
        a=read(),b=read(),c=read(),d=read();
        e[a].push_back((node){b,c,d});
        e[b].push_back((node){a,c,d});
    }
    
	pre(1,0);
    for(int i=1;i<=17;i++)
        for(int j=1;j<=n;j++)fa[j][i]=fa[fa[j][i-1]][i-1];	
        
    for(int i=1,u,v,c,d;i<=m;i++){
       c=read(),d=read(),u=read(),v=read();
       q[u].push_back((ques){i,c,d,1});
       q[v].push_back((ques){i,c,d,1});
       q[LCA(u,v)].push_back((ques){i,c,d,-2});
    }
	dfs(1,0,0);
	
    for(int i=1;i<=m;i++)printf("%d\n",ret[i]);
}