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 M (1000000007)
#define N (100010)
struct data{
	int to,col,length;
};
struct node{
	int id,col,length,cnt;
};
int n,q,all,cnt[N],dep[N],len[N],ans[N],f[N][20];
vector<data>G[N];
vector<node>Q[N];

void dfs(int x,int fa){
	dep[x]=dep[fa]+1; f[x][0]=fa;
	for (int i=0;i<G[x].size();i++){
		int y=G[x][i].to;
		if (y!=fa) dfs(y,x);
	}
}

int LCA(int x,int y){
	if (dep[x]<dep[y]) swap(x,y);
	int step=dep[x]-dep[y];
	for (int i=0;i<=18;i++) 
		if ((1<<i)&step) x=f[x][i];
	if (x==y) return x;
	for (int i=18;i>=0;i--)
		if (f[x][i]!=f[y][i])
			x=f[x][i],y=f[y][i];
	return f[x][0];
}

void query(int x){
	for (int i=0;i<G[x].size();i++){
		data y=G[x][i];
		if (y.to==f[x][0]) continue;
		++cnt[y.col]; len[y.col]+=y.length; all+=y.length;
		query(y.to);
		--cnt[y.col]; len[y.col]-=y.length; all-=y.length;
	}
	for (int i=0;i<Q[x].size();i++){
		node now=Q[x][i];
		ans[now.id]+=now.cnt*(all-len[now.col]+cnt[now.col]*now.length);
	}
}

int main(){
	scanf("%d%d",&n,&q);
	int x,y,col,length;
	for (int i=1;i<n;i++){
		scanf("%d%d%d%d",&x,&y,&col,&length);
		G[x].push_back({y,col,length});
		G[y].push_back({x,col,length});
	}
	dfs(1,0);
	for (int i=1;i<=18;i++)
		for (int j=1;j<=n;j++)
			f[j][i]=f[f[j][i-1]][i-1];
	for (int i=1;i<=q;i++){
		scanf("%d%d%d%d",&col,&length,&x,&y);
		int lca=LCA(x,y);
		Q[x].push_back({i,col,length,1});
		Q[y].push_back({i,col,length,1});
		Q[lca].push_back({i,col,length,-2});
	}
	query(1);
	for (int i=1;i<=q;i++) printf("%d\n",ans[i]);
}