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;
typedef int ll;
struct E{
	int to;
	int clr;
	int len;
};
vector<E> g[100005];
int anc[18][100005],d[100005];
void dfs(int u,int fa){
	for(auto v:g[u]){
		if(v.to==fa){
			continue;
		}
		anc[0][v.to]=u;
		d[v.to]=d[u]+1;
		dfs(v.to,u);
	}
}
int LCA(int x,int y){
	if(d[x]>d[y]){
		swap(x,y);
	}
	for(int ind=0;d[x]<d[y];++ind){
		if((d[y]-d[x])>>ind&1){
			y=anc[ind][y];
		}
	}
	if(x==y){
		return x;
	}
	for(int ind=18;ind--;){
		if(anc[ind][x]!=anc[ind][y]){
			x=anc[ind][x],y=anc[ind][y];
		}
	}
	return anc[0][x];
}
struct info{
	int ind;
	char typ;
	int clr;
	int cnt;
	info(){}
	info(int in,char ty,int cl,int cn){
		ind=in,typ=ty,clr=cl,cnt=cn;
	}
}; 
vector<info> q[100005];
ll sum[100005];
ll cnt[100005];
ll ans[100005];
void DFS(int u,int fa,ll len){
	for(auto now:q[u]){
		if(now.typ=='d'){
			ans[now.ind]+=len*now.cnt;
		}
		else if(now.typ=='s'){
			ans[now.ind]+=sum[now.clr]*now.cnt;
		}
		else{
			ans[now.ind]+=cnt[now.clr]*now.cnt;
		}
	}
	for(auto v:g[u]){
		if(v.to==fa){
			continue;
		}
		sum[v.clr]+=v.len;
		cnt[v.clr]++;
		DFS(v.to,u,len+v.len);
		sum[v.clr]-=v.len;
		cnt[v.clr]--;
	}
}
int main(){
	int n,Q;
	cin>>n>>Q;
	for(int i=1;i<n;++i){
		int u,v,clr,leng;
		cin>>u>>v>>clr>>leng; 
		u--,v--,clr--;
		E a;
		a.to=v,a.len=leng,a.clr=clr;
		E b;
		b.to=u,b.len=leng,b.clr=clr;
		g[u].push_back(a),g[v].push_back(b);
	}
	dfs(0,0);
	for(int ind=0;ind<17;++ind){
		for(int u=0;u<n;++u){
			anc[ind+1][u]=anc[ind][anc[ind][u]];
		}
	}
	for(int i=0;i<Q;++i){
		int clr,leng,u,v;
		cin>>clr>>leng>>u>>v;
		clr--;
		u--;
		v--;
		int lca=LCA(u,v);
		q[u].emplace_back(i,'d',0,1);
		q[u].emplace_back(i,'s',clr,-1);
		q[u].emplace_back(i,'c',clr,leng);
		q[v].emplace_back(i,'d',0,1);
		q[v].emplace_back(i,'s',clr,-1);
		q[v].emplace_back(i,'c',clr,leng);
		q[lca].emplace_back(i,'d',0,-2);
		q[lca].emplace_back(i,'s',clr,2);
		q[lca].emplace_back(i,'c',clr,-2*leng);
	}
	DFS(0,0,0);
	for(int i=0;i<Q;++i){
		cout<<ans[i]<<'\n';
	}
	return 0;
}