#include<bits/stdc++.h>
#define N 100002
using namespace std;
const int p=1e9+7;
struct Que{
	int c,w,id,f;
};
struct node{
	int to,c,w;
};
int n,m,fa[N][19],nc[N],len[N],dis[N],ans[N],sc[N],dep[N];
vector<node>s[N];
vector<Que>q[N];
void dfs1(int x,int f,int w){
	dep[x]=dep[f]+1;fa[x][0]=f;dis[x]=dis[f]+w;
	for(int i=0;i<(int)s[x].size();i++){
		int y=s[x][i].to;
		if(y==f)continue;
		dfs1(y,x,s[x][i].w);
	}
}
void dfs2(int x,int f){
	for(int i=0;i<(int)s[x].size();i++){
		int c=s[x][i].c,y=s[x][i].to,w=s[x][i].w;
		if(y==f)continue;
		nc[c]++;sc[c]+=w;
		dfs2(y,x);
		nc[c]--;sc[c]-=w;
	}
	for(int i=0;i<(int)q[x].size();i++){
		int c=q[x][i].c;
		ans[q[x][i].id]+=q[x][i].f*(dis[x]-sc[c]+nc[c]*q[x][i].w);
	}
}
int Up(int x,int step){
	for(int i=0;i<19;i++){
		if(step&(1<<i))x=fa[x][i];
	}
	return x;
}
int LCA(int x,int y){
	if(dep[x]>dep[y])swap(x,y);
	y=Up(y,dep[y]-dep[x]);
	if(y==x)return y;
	for(int i=18;i>=0;i--){
		if(fa[x][i]!=fa[y][i]){
			x=fa[x][i];
			y=fa[y][i];
		}
	}
	return fa[x][0];
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1,u,v,c,w;i<n;i++){
		scanf("%d%d%d%d",&u,&v,&c,&w);
		s[u].push_back((node)<%v,c,w%>);s[v].push_back((node)<%u,c,w%>);
	}
	dfs1(1,0,0);
	for(int j=1;j<19;j++)for(int i=1;i<=n;i++)fa[i][j]=fa[fa[i][j-1]][j-1];
	for(int i=1,x,y,u,v;i<=m;i++){
		scanf("%d%d%d%d",&x,&y,&u,&v);
		int L=LCA(u,v);
		q[u].push_back((Que)<%x,y,i,1%>);q[v].push_back((Que)<%x,y,i,1%>);
		q[L].push_back((Que)<%x,y,i,-2%>);
	}
	dfs2(1,0);
	for(int i=1;i<=m;i++)printf("%d\n",ans[i]);
	return 0;
}
### Prompt

Please create a solution in cpp 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
#include<bits/stdc++.h>
#define N 100002
using namespace std;
const int p=1e9+7;
struct Que{
	int c,w,id,f;
};
struct node{
	int to,c,w;
};
int n,m,fa[N][19],nc[N],len[N],dis[N],ans[N],sc[N],dep[N];
vector<node>s[N];
vector<Que>q[N];
void dfs1(int x,int f,int w){
	dep[x]=dep[f]+1;fa[x][0]=f;dis[x]=dis[f]+w;
	for(int i=0;i<(int)s[x].size();i++){
		int y=s[x][i].to;
		if(y==f)continue;
		dfs1(y,x,s[x][i].w);
	}
}
void dfs2(int x,int f){
	for(int i=0;i<(int)s[x].size();i++){
		int c=s[x][i].c,y=s[x][i].to,w=s[x][i].w;
		if(y==f)continue;
		nc[c]++;sc[c]+=w;
		dfs2(y,x);
		nc[c]--;sc[c]-=w;
	}
	for(int i=0;i<(int)q[x].size();i++){
		int c=q[x][i].c;
		ans[q[x][i].id]+=q[x][i].f*(dis[x]-sc[c]+nc[c]*q[x][i].w);
	}
}
int Up(int x,int step){
	for(int i=0;i<19;i++){
		if(step&(1<<i))x=fa[x][i];
	}
	return x;
}
int LCA(int x,int y){
	if(dep[x]>dep[y])swap(x,y);
	y=Up(y,dep[y]-dep[x]);
	if(y==x)return y;
	for(int i=18;i>=0;i--){
		if(fa[x][i]!=fa[y][i]){
			x=fa[x][i];
			y=fa[y][i];
		}
	}
	return fa[x][0];
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1,u,v,c,w;i<n;i++){
		scanf("%d%d%d%d",&u,&v,&c,&w);
		s[u].push_back((node)<%v,c,w%>);s[v].push_back((node)<%u,c,w%>);
	}
	dfs1(1,0,0);
	for(int j=1;j<19;j++)for(int i=1;i<=n;i++)fa[i][j]=fa[fa[i][j-1]][j-1];
	for(int i=1,x,y,u,v;i<=m;i++){
		scanf("%d%d%d%d",&x,&y,&u,&v);
		int L=LCA(u,v);
		q[u].push_back((Que)<%x,y,i,1%>);q[v].push_back((Que)<%x,y,i,1%>);
		q[L].push_back((Que)<%x,y,i,-2%>);
	}
	dfs2(1,0);
	for(int i=1;i<=m;i++)printf("%d\n",ans[i]);
	return 0;
}
```