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 vi vector<int>
#define pb push_back
#define pii pair<int,int>
#define piii pair<int,pii>
#define F first
#define S second
#define mp make_pair
using namespace std;
const int N=1e5+5;
int fa[N][25],ans[N],n,Q,dep[N],d[N],cnt[N],sum[N];
vi v[N],col[N],len[N];
vector<piii> q[N];
void DFS(int now,int fat)
{
	fa[now][0]=fat;
	dep[now]=dep[fat]+1;
	for(int i=0;i<v[now].size();i++)
		if(fat!=v[now][i])
			d[v[now][i]]=d[now]+len[now][i],DFS(v[now][i],now);
}
int lca(int x,int y)
{
	if(dep[x]>dep[y])swap(x,y);
	for(int i=0;dep[x]<dep[y];i++)
		if(((dep[y]-dep[x])>>i)&1)
			y=fa[y][i];
	if(x==y)return x;
	for(int i=19;i>=0;i--)
		if(fa[x][i]!=fa[y][i])
			x=fa[x][i],y=fa[y][i];
	return fa[x][0];
}
void dfs(int now,int fat)
{
	for(int i=0;i<q[now].size();i++)
		if(q[now][i].F>0){
			ans[q[now][i].F]+=d[now]+cnt[q[now][i].S.F]*q[now][i].S.S-sum[q[now][i].S.F];
		}else{
			ans[-q[now][i].F]-=2*(d[now]+cnt[q[now][i].S.F]*q[now][i].S.S-sum[q[now][i].S.F]);
		}
	for(int i=0;i<v[now].size();i++)
	{
		int to=v[now][i];
		if(to==fat)continue;
		cnt[col[now][i]]++;
		sum[col[now][i]]+=len[now][i];
		dfs(to,now);///////
		cnt[col[now][i]]--;
		sum[col[now][i]]-=len[now][i];
	}
}
int main()
{
	cin>>n>>Q;
	for(int i=0;i<n-1;i++)
	{
		int a,b,c,d;
		cin>>a>>b>>c>>d;
		v[a].pb(b);
		v[b].pb(a);
		col[a].pb(c);
		col[b].pb(c);
		len[a].pb(d);
		len[b].pb(d);
	}
	DFS(1,0);
	for(int i=1;i<20;i++)
		for(int j=1;j<=n;j++)
			fa[j][i]=fa[fa[j][i-1]][i-1];
	for(int i=1;i<=Q;i++)
	{
		int a,b,x,y;
		cin>>a>>b>>x>>y;
		q[x].pb(mp(i,mp(a,b)));
		q[y].pb(mp(i,mp(a,b)));
		q[lca(x,y)].pb(mp(-i,mp(a,b)));
	}
	dfs(1,0);
	for(int i=1;i<=Q;i++)cout<<ans[i]<<"\n";
	return 0;
}