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 <cstdio>
#include <vector>
using namespace std;
#define Maxn 100000
vector<int> G[Maxn+5];
int fa[30][Maxn+5];
int depth[Maxn+5];
int n,q;
struct Edge{
	int to,color,cost;
};
vector<Edge> edge[Maxn+5];
struct Query{
	int color,cost,t,qid;
};
vector<Query> query[Maxn+5];
int dis[Maxn+5];
int cnt[Maxn+5];
int sum[Maxn+5];
int ans[Maxn+5];
int lca(int u,int v){
	if(depth[u]>depth[v]){
		swap(u,v);
	}
	for(int k=0;k<30;k++){
		if(((depth[v]-depth[u])>>k)&1){
			v=fa[k][v];
		}
	}
	if(u==v){
		return u;
	}
	for(int k=29;k>=0;k--){
		if(fa[k][u]!=fa[k][v]){
			u=fa[k][u];
			v=fa[k][v];
		}
	}
	return fa[0][u];
}
void dfs_1(int v,int p,int d){
	fa[0][v]=p;
	depth[v]=d;
	int to;
	for(int i=0;i<(int)G[v].size();i++){
		to=G[v][i];
		if(to==p){
			continue;
		}
		dfs_1(to,v,d+1);
	}
}
void dfs_2(int now,int from){
	Edge e;
	for(int i=0;i<(int)edge[now].size();i++){
		e=edge[now][i];
		if(e.to==from){
			continue;
		}
		dis[e.to]=dis[now]+e.cost;
		dfs_2(e.to,now);
	}
}
void dfs_3(int now,int from){
	int dx;
	for(int i=0;i<(int)query[now].size();i++){
		Query que=query[now][i];
		dx=que.cost*cnt[que.color]-sum[que.color];
		ans[que.qid]+=dx*que.t;
	}
	Edge e;
	for(int i=0;i<(int)edge[now].size();i++){
		e=edge[now][i];
		if(e.to==from){
			continue;
		}
		cnt[e.color]++;
		sum[e.color]+=e.cost;
		dfs_3(e.to,now);
		cnt[e.color]--;
		sum[e.color]-=e.cost;
	}
}
void init(int V){
	dfs_1(0,-1,0);
	for(int k=0;k+1<30;k++){
		for(int v=0;v<V;v++){
			if(fa[k][v]<0){
				fa[k+1][v]=-1;
			}
			else{
				fa[k+1][v]=fa[k][fa[k][v]];
			}
		}
	}
}
int main(){
	scanf("%d%d",&n,&q);
	int a,b,c,d;
	Edge e;
	for(int i=0;i<n-1;i++){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		a--,b--;
		e.to=b;
		e.color=c;
		e.cost=d;
		edge[a].push_back(e);
		e.to=a;
		edge[b].push_back(e);
		G[a].push_back(b);
		G[b].push_back(a);
	}
	init(n);
	dis[0]=0;
	dfs_2(0,-1);
	int x,y,u,v;
	Query que;
	for(int i=0;i<q;i++){
		scanf("%d%d%d%d",&x,&y,&u,&v);
		u--,v--;
		ans[i]=dis[u]+dis[v]-(dis[lca(u,v)]<<1);
		que.color=x;
		que.cost=y;
		que.t=1;
		que.qid=i;
		query[u].push_back(que);
		query[v].push_back(que);
		que.t=-2;
		query[lca(u,v)].push_back(que);
	}
	dfs_3(0,-1);
	for(int i=0;i<q;i++){
		printf("%d\n",ans[i]);
	}
	return 0;
}