#include<bits/stdc++.h>
#define M 100005
using namespace std;
int dep[M],dis[M],n,q,cnt[M],sz[M],fa[M][20],ans[M];
struct node {
	int id,c,d,f;
};
struct Node {
	int to,c,d;
};
vector<Node>G[M];
vector<node>Q[M];
void dfs(int x,int f,int d) {
	dis[x]=d,fa[x][0]=f;
	for(int i=0; i<G[x].size(); i++) {
		Node now=G[x][i];
		if(now.to==f)continue;
		dep[now.to]=dep[x]+1;
		dfs(now.to,x,d+now.d);
	}
}
void up(int &x,int step) {
	for(int i=0; i<20; i++)if(step&(1<<i))x=fa[x][i];
}
int lca(int x,int y) {
	if(dep[y]<dep[x])swap(x,y);
	up(y,dep[y]-dep[x]);
	if(x==y)return x;
	for(int i=19; i>=0; i--)
		if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];
	return fa[x][0];
}
void redfs(int x,int f) {
	for(int i=0; i<Q[x].size(); i++) {
		node now=Q[x][i];
		int c=now.c,d=now.d;
		ans[now.id]+=now.f*(dis[x]-cnt[c]+sz[c]*d);
	}
	for(int i=0; i<G[x].size(); i++) {
		Node now=G[x][i];
		if(now.to==f)continue;
		sz[now.c]++;
		cnt[now.c]+=now.d;
		redfs(now.to,x);
		sz[now.c]--;
		cnt[now.c]-=now.d;
	}
}
int main() {
	scanf("%d%d",&n,&q);
	for(int i=1; i<n; i++) {
		int x,y,c,d;
		scanf("%d%d%d%d",&x,&y,&c,&d);
		G[x].push_back((Node)<%y,c,d%>);
		G[y].push_back((Node)<%x,c,d%>);
	}
	dfs(1,0,0);
	for(int i=1; i<20; i++)for(int j=1; j<=n; j++)fa[j][i]=fa[fa[j][i-1]][i-1];
	for(int i=1; i<=q; i++) {
		int c,d,x,y;
		scanf("%d%d%d%d",&c,&d,&x,&y);
		int L=lca(x,y);
		Q[x].push_back((node)<%i,c,d,1%>);
		Q[y].push_back((node)<%i,c,d,1%>);
		Q[L].push_back((node)<%i,c,d,-2%>);
	}
	redfs(1,0);
	for(int i=1; i<=q; i++)printf("%d\n",ans[i]);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 M 100005
using namespace std;
int dep[M],dis[M],n,q,cnt[M],sz[M],fa[M][20],ans[M];
struct node {
	int id,c,d,f;
};
struct Node {
	int to,c,d;
};
vector<Node>G[M];
vector<node>Q[M];
void dfs(int x,int f,int d) {
	dis[x]=d,fa[x][0]=f;
	for(int i=0; i<G[x].size(); i++) {
		Node now=G[x][i];
		if(now.to==f)continue;
		dep[now.to]=dep[x]+1;
		dfs(now.to,x,d+now.d);
	}
}
void up(int &x,int step) {
	for(int i=0; i<20; i++)if(step&(1<<i))x=fa[x][i];
}
int lca(int x,int y) {
	if(dep[y]<dep[x])swap(x,y);
	up(y,dep[y]-dep[x]);
	if(x==y)return x;
	for(int i=19; i>=0; i--)
		if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];
	return fa[x][0];
}
void redfs(int x,int f) {
	for(int i=0; i<Q[x].size(); i++) {
		node now=Q[x][i];
		int c=now.c,d=now.d;
		ans[now.id]+=now.f*(dis[x]-cnt[c]+sz[c]*d);
	}
	for(int i=0; i<G[x].size(); i++) {
		Node now=G[x][i];
		if(now.to==f)continue;
		sz[now.c]++;
		cnt[now.c]+=now.d;
		redfs(now.to,x);
		sz[now.c]--;
		cnt[now.c]-=now.d;
	}
}
int main() {
	scanf("%d%d",&n,&q);
	for(int i=1; i<n; i++) {
		int x,y,c,d;
		scanf("%d%d%d%d",&x,&y,&c,&d);
		G[x].push_back((Node)<%y,c,d%>);
		G[y].push_back((Node)<%x,c,d%>);
	}
	dfs(1,0,0);
	for(int i=1; i<20; i++)for(int j=1; j<=n; j++)fa[j][i]=fa[fa[j][i-1]][i-1];
	for(int i=1; i<=q; i++) {
		int c,d,x,y;
		scanf("%d%d%d%d",&c,&d,&x,&y);
		int L=lca(x,y);
		Q[x].push_back((node)<%i,c,d,1%>);
		Q[y].push_back((node)<%i,c,d,1%>);
		Q[L].push_back((node)<%i,c,d,-2%>);
	}
	redfs(1,0);
	for(int i=1; i<=q; i++)printf("%d\n",ans[i]);
	return 0;
}
```