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=1e5+10;
int n,m,t;
typedef long long LL;
struct query{
	int x,y,u,v;
	long long ans;
}Q[N];
vector<int> v[N],b[N];

int cnt,head[N],nxt[N<<1],to[N<<1],col[N<<1],len[N<<1];
void add_edge(int x,int y,int c,int d) {
	nxt[++cnt]=head[x]; to[cnt]=y; col[cnt]=c; len[cnt]=d; head[x]=cnt;
}

int dep[N],f[N][20];
void dfs1(int x,int fa,int d) {
	dep[x]=d;
	for (int i=head[x];i;i=nxt[i]) {
		int y=to[i];
		if (y==fa) continue;
		f[y][0]=x;
		for (int j=1;j<=t;j++) f[y][j]=f[f[y][j-1]][j-1];
		dfs1(y,x,d+1);
	}
}

int LCA(int x,int y) {
	if (dep[x]>dep[y]) swap(x,y);
	for (int i=t;i>=0;i--)
		if (dep[f[y][i]]>=dep[x]) y=f[y][i];
	if (x==y) return x;
	for (int i=t;i>=0;i--)
		if (f[x][i]!=f[y][i]) x=f[x][i],y=f[y][i];
	return f[x][0];		
}

int sumc[N],sumd[N];
void dfs2(int x,int fa,int dis) {
	for (int i=0;i<v[x].size();i++) {
		int id=v[x][i];
		Q[id].ans+=(LL)dis*b[x][i];
		Q[id].ans-=(LL)sumd[Q[id].x]*b[x][i];
		Q[id].ans+=(LL)sumc[Q[id].x]*Q[id].y*b[x][i];
	}
	for (int i=head[x];i;i=nxt[i]) {
		int y=to[i];
		if (y==fa) continue;
		sumc[col[i]]++;
		sumd[col[i]]+=len[i];
		dfs2(y,x,dis+len[i]);
		sumc[col[i]]--;
		sumd[col[i]]-=len[i];
	}
}

int main()
{
	cin>>n>>m;
	t=log2(n)+1;
	for (int i=1;i<n;i++) {
		int x,y,c,d;
		scanf("%d%d%d%d",&x,&y,&c,&d);
		add_edge(x,y,c,d);
		add_edge(y,x,c,d);
	}
	dfs1(1,0,1);
	for (int i=1;i<=m;i++) {
		scanf("%d%d%d%d",&Q[i].x,&Q[i].y,&Q[i].u,&Q[i].v);
		v[Q[i].u].push_back(i); b[Q[i].u].push_back(1);
		v[Q[i].v].push_back(i); b[Q[i].v].push_back(1);
		int lca=LCA(Q[i].u,Q[i].v);
		v[lca].push_back(i); b[lca].push_back(-2);
	}
	dfs2(1,0,0);
	for (int i=1;i<=m;i++) printf("%lld\n",Q[i].ans);
	return 0;	
}