/**********************************************************************************
just to save and look at code later
Izeed's code
https://atcoder.jp/contests/abc133/submissions/6305191
*/
#include<bits/stdc++.h>
using namespace std;
#define M 100005
struct D{
	int t,c,d;
};
struct W{
	int id,c,d,p;
};
vector<D>edge[M];
vector<W>Q[M];
int n,F[M][20],q,dep[M],ans[M],cn[M],cd[M],dis[M];
//정점,부모 노드,쿼리, 깊이, 답, 카운트, 누적합, 루트에서 거리
void dfs(int x,int f,int d){
	F[x][0]=f,dep[x]=dep[f]+1,dis[x]=d;
	for(int i=0;i<edge[x].size();i++){
		int y=edge[x][i].t;
		if(y==f)continue;
		dfs(y,x,d+edge[x][i].d);
	}
}
void up(int &x,int step){
	for(int i=0;i<20;i++)
		if(step&(1<<i))x=F[x][i];
}
int lca(int a,int b){
	if(dep[a]>dep[b])swap(a,b);
	up(b,dep[b]-dep[a]);
	if(a==b)return a;
	for(int i=19;i>=0;i--)
		if(F[a][i]!=F[b][i])a=F[a][i],b=F[b][i];
	return F[a][0];
}
void DFS(int x,int f){
	for(int i=0;i<Q[x].size();i++)ans[Q[x][i].id]+=Q[x][i].p*(dis[x]-cd[Q[x][i].c]+cn[Q[x][i].c]*Q[x][i].d); //
	for(int i=0;i<edge[x].size();i++){
		int y=edge[x][i].t;
		if(y==f)continue;
		cn[edge[x][i].c]++;
		cd[edge[x][i].c]+=edge[x][i].d;
		DFS(y,x);
		cn[edge[x][i].c]--;
		cd[edge[x][i].c]-=edge[x][i].d;
	}
}
int main(){
	scanf("%d%d",&n,&q);
	for(int i=1,a,b,c,d;i<n;i++){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		edge[a].push_back((D){b,c,d});
		edge[b].push_back((D){a,c,d});
	}
	dfs(1,0,0);
	for(int i=1;i<=19;i++)
		for(int j=1;j<=n;j++)F[j][i]=F[F[j][i-1]][i-1];
	for(int i=1;i<=q;i++){
		int u,v,c,d;
		scanf("%d%d%d%d",&c,&d,&u,&v);
		Q[lca(u,v)].push_back((W){i,c,d,-2});
		Q[u].push_back((W){i,c,d,1});
		Q[v].push_back((W){i,c,d,1});
	}
	DFS(1,0);
	for(int i=1;i<=q;i++)printf("%d\n",ans[i]);
	return 0;
}

### Prompt

Your task is to create a cpp solution to the following 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

### Response

```cpp
/**********************************************************************************
just to save and look at code later
Izeed's code
https://atcoder.jp/contests/abc133/submissions/6305191
*/
#include<bits/stdc++.h>
using namespace std;
#define M 100005
struct D{
	int t,c,d;
};
struct W{
	int id,c,d,p;
};
vector<D>edge[M];
vector<W>Q[M];
int n,F[M][20],q,dep[M],ans[M],cn[M],cd[M],dis[M];
//정점,부모 노드,쿼리, 깊이, 답, 카운트, 누적합, 루트에서 거리
void dfs(int x,int f,int d){
	F[x][0]=f,dep[x]=dep[f]+1,dis[x]=d;
	for(int i=0;i<edge[x].size();i++){
		int y=edge[x][i].t;
		if(y==f)continue;
		dfs(y,x,d+edge[x][i].d);
	}
}
void up(int &x,int step){
	for(int i=0;i<20;i++)
		if(step&(1<<i))x=F[x][i];
}
int lca(int a,int b){
	if(dep[a]>dep[b])swap(a,b);
	up(b,dep[b]-dep[a]);
	if(a==b)return a;
	for(int i=19;i>=0;i--)
		if(F[a][i]!=F[b][i])a=F[a][i],b=F[b][i];
	return F[a][0];
}
void DFS(int x,int f){
	for(int i=0;i<Q[x].size();i++)ans[Q[x][i].id]+=Q[x][i].p*(dis[x]-cd[Q[x][i].c]+cn[Q[x][i].c]*Q[x][i].d); //
	for(int i=0;i<edge[x].size();i++){
		int y=edge[x][i].t;
		if(y==f)continue;
		cn[edge[x][i].c]++;
		cd[edge[x][i].c]+=edge[x][i].d;
		DFS(y,x);
		cn[edge[x][i].c]--;
		cd[edge[x][i].c]-=edge[x][i].d;
	}
}
int main(){
	scanf("%d%d",&n,&q);
	for(int i=1,a,b,c,d;i<n;i++){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		edge[a].push_back((D){b,c,d});
		edge[b].push_back((D){a,c,d});
	}
	dfs(1,0,0);
	for(int i=1;i<=19;i++)
		for(int j=1;j<=n;j++)F[j][i]=F[F[j][i-1]][i-1];
	for(int i=1;i<=q;i++){
		int u,v,c,d;
		scanf("%d%d%d%d",&c,&d,&u,&v);
		Q[lca(u,v)].push_back((W){i,c,d,-2});
		Q[u].push_back((W){i,c,d,1});
		Q[v].push_back((W){i,c,d,1});
	}
	DFS(1,0);
	for(int i=1;i<=q;i++)printf("%d\n",ans[i]);
	return 0;
}

```