#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

class LCA{
public:
	vector<vector<pair<int,int>>>E;
	vector<vector<int>>par;
	vector<int>d1;
	vector<long long>d2;
	int MAX_LOG=22;
	int n;
	LCA(){}
	LCA(int N){
		n=N;
		E=vector<vector<pair<int,int>>>(n);
		par=vector<vector<int>>(MAX_LOG,vector<int>(n));
		d1=vector<int>(n);
		d2=vector<long long>(n);
	}
	void add_edge(int a,int b,int c){
		E[a].push_back(make_pair(c,b));
		E[b].push_back(make_pair(c,a));
	}
	void add_edge(int a,int b){
		E[a].push_back(make_pair(1,b));
		E[b].push_back(make_pair(1,a));
	}
private:
	void dfs(int u,int p,int a,long long b){
		par[0][u]=p;d1[u]=a;d2[u]=b;
		for(auto&v:E[u]){
			if(v.second!=p)dfs(v.second,u,a+1,b+v.first);
		}
	}
	bool flag=false;
	void init(){
		dfs(0,-1,0,0);
		for(int i=1;i<MAX_LOG;i++)for(int j=0;j<n;j++){
			if(par[i-1][j]==-1)par[i][j]=-1;
			else par[i][j]=par[i-1][par[i-1][j]];
		}
		flag=true;
	}
public:
	int lca(int u,int v){
		if(!flag)init();
		if(d1[u]>d1[v])swap(u,v);
		for(int i=0;i<MAX_LOG;i++){
			if((d1[v]-d1[u])>>i&1)v=par[i][v];
		}
		if(u==v)return u;
		for(int i=MAX_LOG-1;i>=0;i--){
			if(par[i][u]!=par[i][v]){
				u=par[i][u];
				v=par[i][v];
			}
		}
		return par[0][u];
	}
	long long dist(int u,int v){
		int l=lca(u,v);
		return d2[u]+d2[v]-d2[l]*2;
	}
};


int a[200000],b[200000],c[200000],d[200000];
struct st{
	int a,b,c,id;
};
vector<st>E[200000];

int cnt[200000];
ll sum[200000];
ll all;


int x[200000],y[200000],u[200000],v[200000];
ll ans[200000];

vector<st>query[200000];

void dfs(int v,int p){
	for(auto s:query[v]){
		ll c=all-sum[s.a]+cnt[s.a]*s.b;
		ans[s.id]+=s.c*c;
	}
	for(auto u:E[v]){
		if(u.a==p)continue;
		cnt[u.b]++;
		sum[u.b]+=u.c;
		all+=u.c;
		dfs(u.a,v);
		cnt[u.b]--;
		sum[u.b]-=u.c;
		all-=u.c;
	}
}

int main(){
	int n,q;cin>>n>>q;
	LCA lca(n);
	rep(i,n-1){
		scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
		a[i]--;b[i]--;
		E[a[i]].push_back({b[i],c[i],d[i]});
		E[b[i]].push_back({a[i],c[i],d[i]});
		lca.add_edge(a[i],b[i]);
	}
	rep(i,q){
		scanf("%d%d%d%d",&x[i],&y[i],&u[i],&v[i]);u[i]--;v[i]--;
		query[u[i]].push_back({x[i],y[i],1,i});
		query[v[i]].push_back({x[i],y[i],1,i});
		query[lca.lca(u[i],v[i])].push_back({x[i],y[i],-2,i});
	}
	dfs(0,-1);
	rep(i,q){
		printf("%lld\n",ans[i]);
	}
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

class LCA{
public:
	vector<vector<pair<int,int>>>E;
	vector<vector<int>>par;
	vector<int>d1;
	vector<long long>d2;
	int MAX_LOG=22;
	int n;
	LCA(){}
	LCA(int N){
		n=N;
		E=vector<vector<pair<int,int>>>(n);
		par=vector<vector<int>>(MAX_LOG,vector<int>(n));
		d1=vector<int>(n);
		d2=vector<long long>(n);
	}
	void add_edge(int a,int b,int c){
		E[a].push_back(make_pair(c,b));
		E[b].push_back(make_pair(c,a));
	}
	void add_edge(int a,int b){
		E[a].push_back(make_pair(1,b));
		E[b].push_back(make_pair(1,a));
	}
private:
	void dfs(int u,int p,int a,long long b){
		par[0][u]=p;d1[u]=a;d2[u]=b;
		for(auto&v:E[u]){
			if(v.second!=p)dfs(v.second,u,a+1,b+v.first);
		}
	}
	bool flag=false;
	void init(){
		dfs(0,-1,0,0);
		for(int i=1;i<MAX_LOG;i++)for(int j=0;j<n;j++){
			if(par[i-1][j]==-1)par[i][j]=-1;
			else par[i][j]=par[i-1][par[i-1][j]];
		}
		flag=true;
	}
public:
	int lca(int u,int v){
		if(!flag)init();
		if(d1[u]>d1[v])swap(u,v);
		for(int i=0;i<MAX_LOG;i++){
			if((d1[v]-d1[u])>>i&1)v=par[i][v];
		}
		if(u==v)return u;
		for(int i=MAX_LOG-1;i>=0;i--){
			if(par[i][u]!=par[i][v]){
				u=par[i][u];
				v=par[i][v];
			}
		}
		return par[0][u];
	}
	long long dist(int u,int v){
		int l=lca(u,v);
		return d2[u]+d2[v]-d2[l]*2;
	}
};


int a[200000],b[200000],c[200000],d[200000];
struct st{
	int a,b,c,id;
};
vector<st>E[200000];

int cnt[200000];
ll sum[200000];
ll all;


int x[200000],y[200000],u[200000],v[200000];
ll ans[200000];

vector<st>query[200000];

void dfs(int v,int p){
	for(auto s:query[v]){
		ll c=all-sum[s.a]+cnt[s.a]*s.b;
		ans[s.id]+=s.c*c;
	}
	for(auto u:E[v]){
		if(u.a==p)continue;
		cnt[u.b]++;
		sum[u.b]+=u.c;
		all+=u.c;
		dfs(u.a,v);
		cnt[u.b]--;
		sum[u.b]-=u.c;
		all-=u.c;
	}
}

int main(){
	int n,q;cin>>n>>q;
	LCA lca(n);
	rep(i,n-1){
		scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
		a[i]--;b[i]--;
		E[a[i]].push_back({b[i],c[i],d[i]});
		E[b[i]].push_back({a[i],c[i],d[i]});
		lca.add_edge(a[i],b[i]);
	}
	rep(i,q){
		scanf("%d%d%d%d",&x[i],&y[i],&u[i],&v[i]);u[i]--;v[i]--;
		query[u[i]].push_back({x[i],y[i],1,i});
		query[v[i]].push_back({x[i],y[i],1,i});
		query[lca.lca(u[i],v[i])].push_back({x[i],y[i],-2,i});
	}
	dfs(0,-1);
	rep(i,q){
		printf("%lld\n",ans[i]);
	}
}
```