#include<bits/stdc++.h>
using namespace std;
const int N=100001;
struct node {
	int to,cl,le;
};
struct ques {
	int id,cl,le,fg;
};
int n,q;
int f[N][18],dep[N],dis[N],ans[N],cs[N],cc[N];
vector<node>G[N];
vector<ques>Q[N];
void dfs(int x,int fa,int d) {
	f[x][0]=fa;
	dep[x]=dep[fa]+1;
	dis[x]=d;
	for(unsigned i=0; i<G[x].size(); i++) {
		int y=G[x][i].to;
		if(y==fa)continue;
		dfs(y,x,d+G[x][i].le);
	}
}
int UP(int x,int step) {
	for(int i=0; i<=17; i++)if(step&(1<<i))x=f[x][i];
	return x;
}
int LCA(int a,int b) {
	if(dep[a]>dep[b])swap(a,b);
	b=UP(b,dep[b]-dep[a]);
	if(b==a)return a;
	for(int i=17; 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) {
	for(unsigned i=0; i<Q[x].size(); i++) {
		ans[Q[x][i].id]+=Q[x][i].fg*(dis[x]-cs[Q[x][i].cl]+cc[Q[x][i].cl]*Q[x][i].le);
	}
	for(unsigned i=0; i<G[x].size(); i++) {
		int y=G[x][i].to;
		if(y==f[x][0])continue;
		cc[G[x][i].cl]++;
		cs[G[x][i].cl]+=G[x][i].le;
		DFS(y);
		cc[G[x][i].cl]--;
		cs[G[x][i].cl]-=G[x][i].le;
	}
}
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);
		G[a].push_back((node)<%b,c,d%>);
		G[b].push_back((node)<%a,c,d%>);
	}
	dfs(1,0,0);
	for(int j=1; j<18; j++)for(int i=1; i<=n; i++)f[i][j]=f[f[i][j-1]][j-1];
	for(int i=1,a,b,c,d; i<=q; i++) {
		scanf("%d%d%d%d",&a,&b,&c,&d);
		int t=LCA(c,d);
		Q[c].push_back((ques)<%i,a,b,1%>);
		Q[d].push_back((ques)<%i,a,b,1%>);
		Q[t].push_back((ques)<%i,a,b,-2%>);
	}
	DFS(1);
	for(int i=1; i<=q; i++)printf("%d\n",ans[i]);
}
### Prompt

In CPP, your task is to solve 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>
using namespace std;
const int N=100001;
struct node {
	int to,cl,le;
};
struct ques {
	int id,cl,le,fg;
};
int n,q;
int f[N][18],dep[N],dis[N],ans[N],cs[N],cc[N];
vector<node>G[N];
vector<ques>Q[N];
void dfs(int x,int fa,int d) {
	f[x][0]=fa;
	dep[x]=dep[fa]+1;
	dis[x]=d;
	for(unsigned i=0; i<G[x].size(); i++) {
		int y=G[x][i].to;
		if(y==fa)continue;
		dfs(y,x,d+G[x][i].le);
	}
}
int UP(int x,int step) {
	for(int i=0; i<=17; i++)if(step&(1<<i))x=f[x][i];
	return x;
}
int LCA(int a,int b) {
	if(dep[a]>dep[b])swap(a,b);
	b=UP(b,dep[b]-dep[a]);
	if(b==a)return a;
	for(int i=17; 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) {
	for(unsigned i=0; i<Q[x].size(); i++) {
		ans[Q[x][i].id]+=Q[x][i].fg*(dis[x]-cs[Q[x][i].cl]+cc[Q[x][i].cl]*Q[x][i].le);
	}
	for(unsigned i=0; i<G[x].size(); i++) {
		int y=G[x][i].to;
		if(y==f[x][0])continue;
		cc[G[x][i].cl]++;
		cs[G[x][i].cl]+=G[x][i].le;
		DFS(y);
		cc[G[x][i].cl]--;
		cs[G[x][i].cl]-=G[x][i].le;
	}
}
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);
		G[a].push_back((node)<%b,c,d%>);
		G[b].push_back((node)<%a,c,d%>);
	}
	dfs(1,0,0);
	for(int j=1; j<18; j++)for(int i=1; i<=n; i++)f[i][j]=f[f[i][j-1]][j-1];
	for(int i=1,a,b,c,d; i<=q; i++) {
		scanf("%d%d%d%d",&a,&b,&c,&d);
		int t=LCA(c,d);
		Q[c].push_back((ques)<%i,a,b,1%>);
		Q[d].push_back((ques)<%i,a,b,1%>);
		Q[t].push_back((ques)<%i,a,b,-2%>);
	}
	DFS(1);
	for(int i=1; i<=q; i++)printf("%d\n",ans[i]);
}
```