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 <algorithm>
#include <vector>
#include <cstring>
#include <cstdio>
#define int long long
#define pb push_back
#define mp make_pair
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
const int mx=100010;
struct Node{
	int u,co,l;
};
struct Node2{
	int id,co,d,k;
};
int n,m;
int fa[mx][20],dep[mx];
int dis[mx],cnt[mx],ret[mx];
vector<Node>G[mx];
vector<Node2>que[mx];
void pre(int x,int pr){
	fa[x][0]=pr;
	dep[x]=dep[pr]+1;
	for(auto to:G[x])if(to.u!=pr){
		pre(to.u,x);
	}
}
int update(int x,int dep){
	for(int i=0;i<=17;i++)if(dep&(1<<i))x=fa[x][i];
	return x;
}
int lca(int a,int b){
	if(dep[a]>dep[b])swap(a,b);
	b=update(b,dep[b]-dep[a]);
	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];
}
void dfs(int x,int pr,int d){
	for(auto cur:que[x]){
		ret[cur.id]+=cur.k*(d-dis[cur.co]+cnt[cur.co]*cur.d);
	}
	for(auto to:G[x])if(to.u!=pr){
		cnt[to.co]++,dis[to.co]+=to.l;
		dfs(to.u,x,d+to.l);
		cnt[to.co]--,dis[to.co]-=to.l;
	}
}
signed main(){
	cin>>n>>m;
	for(int i=1;i<n;i++){
		int a,b,c,d;
		cin>>a>>b>>c>>d;
		G[a].pb((Node){b,c,d});
		G[b].pb((Node){a,c,d});
	}
	pre(1,0);
	for(int i=1;i<=17;i++)
		for(int j=1;j<=n;j++)
			fa[j][i]=fa[fa[j][i-1]][i-1];
	for(int i=1;i<=m;i++){
		int u,v,c,d;
		cin>>c>>d>>u>>v;
		que[u].pb((Node2){i,c,d,1});
		que[v].pb((Node2){i,c,d,1});
		que[lca(u,v)].pb((Node2){i,c,d,-2});
	}
	dfs(1,0,0);
	for(int i=1;i<=m;i++){
		cout<<ret[i]<<endl;
	}
	return 0;
}