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=100005;
struct edge
{
	int v,x,y;
	edge(){}
	edge(int _v,int _x,int _y):v(_v),x(_x),y(_y){}
};
struct query
{
	int d,u,v,c;
	query(){}
	query(int _d,int _u,int _v,int _c):d(_d),u(_u),v(_v),c(_c){}
};
vector<edge>e[N];
vector<query>q[N];
vector<int>g[N];
int n,m,s[N],t[N],tim,dis[N],fa[N][17],dep[N],sum[N],ans[N],df[N];
void dfs(int u)
{
	s[u]=++tim;
	for(auto i:e[u])
	{
		int v=i.v,x=i.x,y=i.y;
		if(v==fa[u][0])
			continue;
		fa[v][0]=u;
		dep[v]=dep[u]+1;
		dis[v]=dis[u]+y;
		df[v]=y;
		g[x].push_back(v);
		dfs(v);
	}
	t[u]=tim;
}
int lca(int u,int v)
{
	if(dep[u]<dep[v])
		swap(u,v);
	for(int i=16;i>=0;i--)
		if(dep[fa[u][i]]>=dep[v])
			u=fa[u][i];
	if(u==v)
		return u;
	for(int i=16;i>=0;i--)
		if(fa[u][i]!=fa[v][i])
			u=fa[u][i],v=fa[v][i];
	return fa[u][0];
}
struct seg
{
	int tr[N<<2],tag[N<<2];
	void pushdown(int k)
	{
		tr[k<<1]+=tag[k];
		tag[k<<1]+=tag[k];
		tr[k<<1|1]+=tag[k];
		tag[k<<1|1]+=tag[k];
		tag[k]=0;
	}
	void update(int k,int l,int r,int a,int b,int v)
	{
		if(l==a&&r==b)
		{
			tr[k]+=v;
			tag[k]+=v;
			return;
		}
		pushdown(k);
		int mid=l+r>>1;
		if(b<=mid) update(k<<1,l,mid,a,b,v);
		else if(a>mid) update(k<<1|1,mid+1,r,a,b,v);
		else update(k<<1,l,mid,a,mid,v),update(k<<1|1,mid+1,r,mid+1,b,v);
	}
	int query(int k,int l,int r,int x)
	{
		if(l==r)
			return tr[k];
		pushdown(k);
		int mid=l+r>>1;
		if(x<=mid) return query(k<<1,l,mid,x);
		else return query(k<<1|1,mid+1,r,x);
	}
}s1,s2;
int dist(int u,int x,int y)
{
	return dis[u]-s2.query(1,1,n,s[u])+s1.query(1,1,n,s[u])*y;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<n;i++)
	{
		int u,v,x,y;
		scanf("%d%d%d%d",&u,&v,&x,&y);
		e[u].emplace_back(v,x,y);
		e[v].emplace_back(u,x,y);
		sum[x]+=y;
	}
	for(int i=1;i<=m;i++)
	{
		int c,d,u,v;
		scanf("%d%d%d%d",&c,&d,&u,&v);
		q[c].emplace_back(d,u,v,i);
	}
	dep[1]=1;
	dfs(1);
	for(int j=1;j<17;j++)
		for(int i=1;i<=n;i++)
			fa[i][j]=fa[fa[i][j-1]][j-1];
	for(int i=1;i<=n;i++)
	{
		for(auto u:g[i])
		{
			s1.update(1,1,n,s[u],t[u],1);
			s2.update(1,1,n,s[u],t[u],df[u]);
		}
		for(auto j:q[i])
		{
			int c=j.c,u=j.u,v=j.v,l=lca(u,v),y=j.d;
			ans[c]=dist(u,i,y)+dist(v,i,y)-2*dist(l,i,y);
		}
		for(auto u:g[i])
		{
			s1.update(1,1,n,s[u],t[u],-1);
			s2.update(1,1,n,s[u],t[u],-df[u]);
		}
	}
	for(int i=1;i<=m;i++)
		printf("%d\n",ans[i]);
	return 0;
}