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<iostream>
#include<stdio.h>
#include<vector>
using namespace std;

const int maxn=1e5+5;
int n;
struct node {
	int to,c,v;
};

struct query {
	int id,c,d,flag;
};

vector<query>q[maxn];
vector<node>a[maxn];
int p[maxn][20],dep[maxn],dis[maxn],s[maxn],t[maxn],ans[maxn];

void dfs(int x,int f,int d,int s) {
	p[x][0]=f,dep[x]=d,dis[x]=s;
	for(int i=0; i<(int)a[x].size(); i++) {
		int y=a[x][i].to;
		if(y==f)continue;
		dfs(y,x,d+1,s+a[x][i].v);
	}
}

void init() {
	for(int j=1; j<20; j++)
		for(int i=1; i<=n; i++)
			p[i][j]=p[p[i][j-1]][j-1];
}

int up(int x,int step) {
	for(int i=0; i<20; i++)
		if((1<<i)&step)x=p[x][i];
	return x;
}

int LCA(int a,int b) {
//	puts("1");
	if(dep[a]>dep[b])swap(a,b);
	b=up(b,dep[b]-dep[a]);
	if(a==b)return a;
	for(int i=19; i>=0; i--)
		if(p[a][i]!=p[b][i])
			a=p[a][i],b=p[b][i];
	return p[a][0];
}

void DFS(int x,int f) {
	for(int i=0; i<(int)q[x].size(); i++) {
		int c=q[x][i].c;
		int d=q[x][i].d;
		ans[q[x][i].id]+=q[x][i].flag*(dis[x]-t[c]+s[c]*d);
	}
	for(int i=0; i<(int)a[x].size(); i++) {
		int y=a[x][i].to;
		if(y==f)continue;
		s[a[x][i].c]++;
		t[a[x][i].c]+=a[x][i].v;
		DFS(y,x);
		s[a[x][i].c]--;
		t[a[x][i].c]-=a[x][i].v;
	}
}

int main() {
	int m;
	scanf("%d %d",&n,&m);
	for(int i=1; i<n; i++) {
		int x,y,c,d;
		scanf("%d %d %d %d",&x,&y,&c,&d);
		a[x].push_back((node)<%y,c,d%>);
		a[y].push_back((node)<%x,c,d%>);
	}
	dfs(1,0,0,0);
	init();
	for(int i=1; i<=m; i++) {
		int x,y,c,d;
		scanf("%d %d %d %d",&c,&d,&x,&y);
		q[x].push_back((query)<%i,c,d,1%>);
		q[y].push_back((query)<%i,c,d,1%>);
		q[LCA(x,y)].push_back((query)<%i,c,d,-2%>);
	}
//	puts("1");
	DFS(1,0);
	for(int i=1; i<=m; i++)printf("%d\n",ans[i]);
	return 0;
}