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>
#define LL long long
#define N 100005
#define M 20*N
#define db(x) cout<<#x<<"="<<x<<endl;
using namespace std;
struct node{
	int cnt;
	LL sum;
};
struct edge {
	int to,nxt,c,d;
} e[N<<1];
int n,q,head[N],dis[N];
int rt[N],tot;
int lson[M],rson[M],cnt[M];
LL sum[M];
int fa[N][18],dep[N];
void link(int s,int t,int c,int d,int n) {
	e[n]=(edge)<%t,head[s],c,d%>;
	head[s]=n;
}
void insert(int &nt,int ot,int l,int r,int x,int k) {
	nt=++tot;
	lson[nt]=lson[ot];
	rson[nt]=rson[ot];
	cnt[nt]=cnt[ot]+1;
	sum[nt]=sum[ot]+k;
	if(l==r)return;
	int mid=l+r>>1;
	if(x<=mid)insert(lson[nt],lson[ot],l,mid,x,k);
	else insert(rson[nt],rson[ot],mid+1,r,x,k);
}
node query(int u,int v,int lca,int l,int r,int x){
	if(l==r)return (node)<%cnt[u]+cnt[v]-2*cnt[lca],sum[u]+sum[v]-2*sum[lca]%>;
	int mid=l+r>>1;
	if(x<=mid)return query(lson[u],lson[v],lson[lca],l,mid,x);
	else return query(rson[u],rson[v],rson[lca],mid+1,r,x);
}
void dfs(int x,int f,int c,int d) {
	fa[x][0]=f,dep[x]=dep[f]+1;
	if(x!=1)dis[x]=dis[f]+d;	
	if(x!=1)insert(rt[x],rt[f],1,n,c,d);
	for(int i=head[x]; i; i=e[i].nxt) {
		int to=e[i].to;
		if(to==f)continue;
		dfs(to,x,e[i].c,e[i].d);
	}
}
int Lca(int a,int b) {
	if(dep[a]>dep[b])swap(a,b);
	int s=dep[b]-dep[a];
	for(int i=0; 1<<i<=n; i++)if(s&(1<<i))b=fa[b][i];
	if(a==b)return a;
	for(int i=17; i>=0; i--)if(fa[a][i]!=fa[b][i])a=fa[a][i],b=fa[b][i];
	return fa[a][0];
}
int main() {
	scanf("%d%d",&n,&q);
	for(int i=1; i<n; i++) {
		int s,t,c,d;
		scanf("%d%d%d%d",&s,&t,&c,&d);
		link(s,t,c,d,(i<<1)-1);
		link(t,s,c,d,i<<1);
	}
	dfs(1,0,0,0);
	for(int i=1; 1<<i<=n; i++)
		for(int j=1; j<=n; j++)fa[j][i]=fa[fa[j][i-1]][i-1];
	while(q--) {
		int x,y,u,v;
		scanf("%d%d%d%d",&x,&y,&u,&v);
		int lca=Lca(u,v);
		node now=query(rt[u],rt[v],rt[lca],1,n,x);
		printf("%lld\n",dis[u]+dis[v]-2*dis[lca]+(1ll*now.cnt*y-now.sum));
	}
	return 0;
}