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;
#define M 200005
struct hh{
	int p,w,id;
};
vector<hh>d[M];
int sz[M],son[M],fa[M],tp[M],dp[M],rt[M],ls[M*20],rs[M*20],sum[M*20],n,cnt,Sz[M*20],dis[M];
long long ans;
void update(int rt1,int &rt2,int l,int r,int x,int w){
	rt2=++cnt;ls[rt2]=ls[rt1];rs[rt2]=rs[rt1];sum[rt2]=sum[rt1]+w;Sz[rt2]=Sz[rt1]+1;
	if(l==r)return;
	int mid=(l+r)>>1;
	if(x<=mid)update(ls[rt1],ls[rt2],l,mid,x,w);
	else update(rs[rt1],rs[rt2],mid+1,r,x,w);
}
void dfs(int x,int f,int w,int p){
	sz[x]=1;dp[x]=dp[f]+1;fa[x]=f;dis[x]=dis[f]+w;
	if(w)update(rt[f],rt[x],1,n,p,w);
	for(int i=0;i<(int)d[x].size();++i){
		int v=d[x][i].p;
		if(v==f)continue;
		dfs(v,x,d[x][i].w,d[x][i].id);sz[x]+=sz[v];
		if(sz[v]>sz[son[x]])son[x]=v;
	}
}
void Dfs(int x,int top){
	tp[x]=top;
	if(son[x])Dfs(son[x],top);
	for(int i=0;i<(int)d[x].size();++i){
		int v=d[x][i].p;
		if(v==fa[x]||v==son[x])continue;
		Dfs(v,v);
	}
}
int Lca(int x,int y){
	while(tp[x]!=tp[y]){
		if(dp[tp[x]]>dp[tp[y]])x=fa[tp[x]];
		else y=fa[tp[y]];
	}
	return dp[x]>dp[y]?y:x;
}
void query(int rt1,int rt2,int rt3,int l,int r,int x,int w){
	if(l==r){
		ans-=sum[rt1]+sum[rt2]-sum[rt3]*2;
		ans+=(Sz[rt1]+Sz[rt2]-Sz[rt3]*2)*w;
		return;
	}
	int mid=(l+r)>>1;
	if(x<=mid)query(ls[rt1],ls[rt2],ls[rt3],l,mid,x,w);
	else query(rs[rt1],rs[rt2],rs[rt3],mid+1,r,x,w);
}
int main(){
	int q;
	scanf("%d%d",&n,&q);
	for(int i=1,a,b,c,f;i<n;++i){
		scanf("%d%d%d%d",&a,&b,&c,&f);
		d[a].push_back((hh){b,f,c});
		d[b].push_back((hh){a,f,c});
	}
	dfs(1,0,0,1);
	Dfs(1,1);
	for(int i=1,x,y,u,v;i<=q;++i){
		scanf("%d%d%d%d",&x,&y,&u,&v);
		int lca=Lca(u,v);ans=dis[u]-2*dis[lca]+dis[v];
		query(rt[u],rt[v],rt[lca],1,n,x,y);
		printf("%lld\n",ans);
	}
}