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 LL long long
#define Mod 1000000007
#define M 100005
int n,m,a,b,c,d;
int Dis[M],Cnt[M];
int Ans[M],Deep[M];
int f[M][20];
struct Node {
	int id,c,a,k;
};
struct Edge {
	int x,c,d;
};
vector<Edge>E[M];
vector<Node>Q[M];
int Up(int x,int Step) {
	for(int i=0; i<20; i++)
		if(Step&(1<<i))x=f[x][i];
	return x;
}
int LCA(int a,int b) {
	if(Deep[a]>Deep[b])swap(a,b);
	b=Up(b,Deep[b]-Deep[a]);
	if(a==b)return a;
	for(int i=19; ~i; --i)
		if(f[a][i]!=f[b][i]) {
			a=f[a][i];
			b=f[b][i];
		}
	return f[a][0];
}
void pre_dfs(int x,int F) {
	f[x][0]=F,Deep[x]=Deep[F]+1;
	for(int i=1; i<20; i++)
		f[x][i]=f[f[x][i-1]][i-1];
	for(int i=0; i<E[x].size(); i++) {
		int u=E[x][i].x;
		if(u==F)continue;
		pre_dfs(u,x);
	}
}
void dfs(int x,int d) {
	for(int i=0; i<Q[x].size(); i++) {
		Node T=Q[x][i];
		Ans[T.id]+=T.k*(d-Dis[T.c]+Cnt[T.c]*T.a);
	}
	for(int i=0; i<E[x].size(); i++) {
		Edge T=E[x][i];
		if(T.x==f[x][0])continue;
		++Cnt[T.c];
		Dis[T.c]+=T.d;
		dfs(T.x,T.d+d);
		--Cnt[T.c];
		Dis[T.c]-=T.d;
	}
}
int main() {
	scanf("%d%d",&n,&m);
	for(int i=1; i<n; i++) {
		scanf("%d%d%d%d",&a,&b,&c,&d);
		E[a].push_back((Edge)<%b,c,d%>);
		E[b].push_back((Edge)<%a,c,d%>);
	}
	pre_dfs(1,0);
	for(int i=1; i<=m; i++) {
		scanf("%d%d%d%d",&c,&d,&a,&b);
		int lca=LCA(a,b);
		Q[lca].push_back((Node)<%i,c,d,-2%>);
		Q[a].push_back((Node)<%i,c,d,1%>);
		Q[b].push_back((Node)<%i,c,d,1%>);
	}
	dfs(1,0);
	for(int i=1; i<=m; i++)printf("%d\n",Ans[i]);
}