#include<bits/stdc++.h>
#define LL long long
#define M 100005
using namespace std;
int n,Q,h[M],tt;
struct edge{
	int nxt,to,c;
	LL d;
}G[M<<1];
void Add(int a,int b,int c,LL d){
	G[++tt]=(edge){h[a],b,c,d};
	h[a]=tt;
}
LL ans[M],cnt[M],ds[M],dep[M];
struct que{int id,x,y,f;};
vector<que>wk[M];
int fa[M][20];
int LCA(int a,int b){
	if(dep[a]>dep[b])swap(a,b);
	int step=dep[b]-dep[a];
	for(int i=19;i>=0;i--)
		if(step&1<<i)b=fa[b][i];
	if(a==b)return a;
	for(int i=19;i>=0;i--)
		if(fa[a][i]!=fa[b][i])a=fa[a][i],b=fa[b][i];
	return fa[a][0];
} 
void dfs_init(int x,int f,int d){
	fa[x][0]=f;dep[x]=d;
	for(int i=h[x];i;i=G[i].nxt){
		int u=G[i].to;
		if(u==f)continue;
		dfs_init(u,x,d+1);
	}
}
void dfs(int x,int f,LL dd){
	for(int i=0;i<wk[x].size();i++){
		int id=wk[x][i].id,a=wk[x][i].x,b=wk[x][i].y,ff=wk[x][i].f;
		ans[id]+=1LL*ff*(dd-ds[a]+cnt[a]*b);
	}
	for(int i=h[x];i;i=G[i].nxt){
		int u=G[i].to,c=G[i].c,d=G[i].d;
		if(u==f)continue;
		cnt[c]++;ds[c]+=d;
		dfs(u,x,dd+d);
		cnt[c]--;ds[c]-=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);
		Add(a,b,c,d);Add(b,a,c,d);
	}
	dfs_init(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<=Q;i++){
		scanf("%d%d%d%d",&x,&y,&u,&v);
		wk[u].push_back((que){i,x,y,1});		
		wk[v].push_back((que){i,x,y,1});
		int lca=LCA(u,v);
		wk[lca].push_back((que){i,x,y,-2});
	}
	dfs(1,0,0);
	for(int i=1;i<=Q;i++)
		printf("%lld\n",ans[i]);
	return 0;
}
### Prompt

Please formulate 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
#include<bits/stdc++.h>
#define LL long long
#define M 100005
using namespace std;
int n,Q,h[M],tt;
struct edge{
	int nxt,to,c;
	LL d;
}G[M<<1];
void Add(int a,int b,int c,LL d){
	G[++tt]=(edge){h[a],b,c,d};
	h[a]=tt;
}
LL ans[M],cnt[M],ds[M],dep[M];
struct que{int id,x,y,f;};
vector<que>wk[M];
int fa[M][20];
int LCA(int a,int b){
	if(dep[a]>dep[b])swap(a,b);
	int step=dep[b]-dep[a];
	for(int i=19;i>=0;i--)
		if(step&1<<i)b=fa[b][i];
	if(a==b)return a;
	for(int i=19;i>=0;i--)
		if(fa[a][i]!=fa[b][i])a=fa[a][i],b=fa[b][i];
	return fa[a][0];
} 
void dfs_init(int x,int f,int d){
	fa[x][0]=f;dep[x]=d;
	for(int i=h[x];i;i=G[i].nxt){
		int u=G[i].to;
		if(u==f)continue;
		dfs_init(u,x,d+1);
	}
}
void dfs(int x,int f,LL dd){
	for(int i=0;i<wk[x].size();i++){
		int id=wk[x][i].id,a=wk[x][i].x,b=wk[x][i].y,ff=wk[x][i].f;
		ans[id]+=1LL*ff*(dd-ds[a]+cnt[a]*b);
	}
	for(int i=h[x];i;i=G[i].nxt){
		int u=G[i].to,c=G[i].c,d=G[i].d;
		if(u==f)continue;
		cnt[c]++;ds[c]+=d;
		dfs(u,x,dd+d);
		cnt[c]--;ds[c]-=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);
		Add(a,b,c,d);Add(b,a,c,d);
	}
	dfs_init(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<=Q;i++){
		scanf("%d%d%d%d",&x,&y,&u,&v);
		wk[u].push_back((que){i,x,y,1});		
		wk[v].push_back((que){i,x,y,1});
		int lca=LCA(u,v);
		wk[lca].push_back((que){i,x,y,-2});
	}
	dfs(1,0,0);
	for(int i=1;i<=Q;i++)
		printf("%lld\n",ans[i]);
	return 0;
}
```