#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

const int maxn=100003;

int dfn[maxn],pre[maxn][23],root[maxn],num[maxn*23],N,dep[maxn],head[maxn],n,Q,ls[maxn*23],rs[maxn*23];
ll sum[maxn*23],dis[maxn];

struct EDGE{
	int to,nxt,w,c;
}edge[maxn<<1];

void addedge(int u,int v,int c,int w){
	static int sz=0;edge[++sz].to=v;edge[sz].nxt=head[u];
	edge[sz].w=w;edge[sz].c=c;head[u]=sz;
}

int newnode(int x){
	++N;
	ls[N]=ls[x];
	rs[N]=rs[x];
	sum[N]=sum[x];
	num[N]=num[x];
	return N;
}

void update(int id,int l,int r,int pos,int w){
	++num[id];
	sum[id]+=w;
	if(l==r) return ;
	int mid=(l+r)/2;
	if(pos<=mid){
		ls[id]=newnode(ls[id]);
		update(ls[id],l,mid,pos,w);
	}else{
		rs[id]=newnode(rs[id]);
		update(rs[id],mid+1,r,pos,w);
	}
}

pair<int,ll> query(int id,int l,int r,int pos){
	if(l==r)return make_pair(num[id],sum[id]);
	int mid=(l+r)/2;
	if(pos<=mid)return query(ls[id],l,mid,pos);
	else return query(rs[id],mid+1,r,pos);
}

void dfs(int u){
	int to;
	dep[u]=dep[pre[u][0]]+1;
	for(int i=head[u];i;i=edge[i].nxt){
		to=edge[i].to;
		if(to==pre[u][0]) continue;
		root[to]=newnode(root[u]);
		update(root[to],1,n,edge[i].c,edge[i].w);
		pre[to][0]=u;
		for(int j=1;j<23;++j)
			pre[to][j]=pre[pre[to][j-1]][j-1];
		dis[to]=dis[u]+edge[i].w;
		dfs(to);
	}
}

int lca(int u,int v){
	if(dep[u]<dep[v])swap(u,v);
	for(int i=20;i>=0;--i)
		if(dep[pre[u][i]]>=dep[v])
			u=pre[u][i];
	if(u==v)return u;
	for(int i=20;i>=0;--i)
		if(pre[u][i]!=pre[v][i])
			u=pre[u][i],v=pre[v][i];
	return pre[u][0];
}

ll D(int u,int x,int y){
	pair<int,ll> p=query(root[u],1,n,x);
//	printf("%d : %d %lld %lld %d !\n",u,p.first,p.second,dis[u],x);
	return dis[u]-p.second+1ll*y*p.first;
}

int main(){
	scanf("%d%d",&n,&Q);
	for(int i=1;i<n;++i){
		int u,v,c,w;
		scanf("%d%d%d%d",&u,&v,&c,&w);
		addedge(u,v,c,w);
		addedge(v,u,c,w);
	}
	root[1]=++N;
	dfs(1);
//	for(int i=1;i<=n;++i)
//		printf("%lld ",dis[i]);
//	puts("");
//	for(int i=1;i<=n;++i)
//		printf("%d ",root[i]);
//	puts("");
	while(Q--){
		int x,y,u,v;
		scanf("%d%d%d%d",&x,&y,&u,&v);
		printf("%lld\n",D(u,x,y)+D(v,x,y)-2*D(lca(u,v),x,y));
	}
	return 0;
}


### Prompt

Create a solution in cpp for 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;

typedef long long ll;
typedef pair<int,int> pii;

const int maxn=100003;

int dfn[maxn],pre[maxn][23],root[maxn],num[maxn*23],N,dep[maxn],head[maxn],n,Q,ls[maxn*23],rs[maxn*23];
ll sum[maxn*23],dis[maxn];

struct EDGE{
	int to,nxt,w,c;
}edge[maxn<<1];

void addedge(int u,int v,int c,int w){
	static int sz=0;edge[++sz].to=v;edge[sz].nxt=head[u];
	edge[sz].w=w;edge[sz].c=c;head[u]=sz;
}

int newnode(int x){
	++N;
	ls[N]=ls[x];
	rs[N]=rs[x];
	sum[N]=sum[x];
	num[N]=num[x];
	return N;
}

void update(int id,int l,int r,int pos,int w){
	++num[id];
	sum[id]+=w;
	if(l==r) return ;
	int mid=(l+r)/2;
	if(pos<=mid){
		ls[id]=newnode(ls[id]);
		update(ls[id],l,mid,pos,w);
	}else{
		rs[id]=newnode(rs[id]);
		update(rs[id],mid+1,r,pos,w);
	}
}

pair<int,ll> query(int id,int l,int r,int pos){
	if(l==r)return make_pair(num[id],sum[id]);
	int mid=(l+r)/2;
	if(pos<=mid)return query(ls[id],l,mid,pos);
	else return query(rs[id],mid+1,r,pos);
}

void dfs(int u){
	int to;
	dep[u]=dep[pre[u][0]]+1;
	for(int i=head[u];i;i=edge[i].nxt){
		to=edge[i].to;
		if(to==pre[u][0]) continue;
		root[to]=newnode(root[u]);
		update(root[to],1,n,edge[i].c,edge[i].w);
		pre[to][0]=u;
		for(int j=1;j<23;++j)
			pre[to][j]=pre[pre[to][j-1]][j-1];
		dis[to]=dis[u]+edge[i].w;
		dfs(to);
	}
}

int lca(int u,int v){
	if(dep[u]<dep[v])swap(u,v);
	for(int i=20;i>=0;--i)
		if(dep[pre[u][i]]>=dep[v])
			u=pre[u][i];
	if(u==v)return u;
	for(int i=20;i>=0;--i)
		if(pre[u][i]!=pre[v][i])
			u=pre[u][i],v=pre[v][i];
	return pre[u][0];
}

ll D(int u,int x,int y){
	pair<int,ll> p=query(root[u],1,n,x);
//	printf("%d : %d %lld %lld %d !\n",u,p.first,p.second,dis[u],x);
	return dis[u]-p.second+1ll*y*p.first;
}

int main(){
	scanf("%d%d",&n,&Q);
	for(int i=1;i<n;++i){
		int u,v,c,w;
		scanf("%d%d%d%d",&u,&v,&c,&w);
		addedge(u,v,c,w);
		addedge(v,u,c,w);
	}
	root[1]=++N;
	dfs(1);
//	for(int i=1;i<=n;++i)
//		printf("%lld ",dis[i]);
//	puts("");
//	for(int i=1;i<=n;++i)
//		printf("%d ",root[i]);
//	puts("");
	while(Q--){
		int x,y,u,v;
		scanf("%d%d%d%d",&x,&y,&u,&v);
		printf("%lld\n",D(u,x,y)+D(v,x,y)-2*D(lca(u,v),x,y));
	}
	return 0;
}


```