#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<int,ll> pil;
typedef pair<ll,int> pli;
#define rep(i,n) for (int i=0;i<n;++i)
#define REP(i,n) for (int i=1;i<=n;++i)
#define all(x) x.begin(),x.end()
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define read(x) scanf("%d",&x)
int n,Q;
vector<pii> G[100005];
int col[100005],len[100005];
int dep[100005];
int anc[100005][20];
inline void dfs(int u,int pa=0){
	dep[u]=dep[pa]+1;
	anc[u][0]=pa;
	for (int i=0;i<G[u].size();++i){
		int v=G[u][i].F;
		if (v!=pa) dfs(v,u);
	}
}
inline int LCA(int x,int y){
	if (dep[x]<dep[y]) swap(x,y);
	int d=dep[x]-dep[y];
	rep(j,19) if (d&(1<<j)) x=anc[x][j];
	if (x==y) return x;
	for (int j=18;j>=0;--j) if (anc[x][j]!=anc[y][j]) x=anc[x][j],y=anc[y][j];
	return anc[x][0];
}
ll ans[100005];
vector<pair<int,pii> > contrib[100005];// F:query id    S: F:col(-:double minus) S: y
ll cnt[100005],sum[100005],dist;
inline void calc(int u,int pa){
	for (int i=0;i<contrib[u].size();++i){
		int id=contrib[u][i].F,c=contrib[u][i].S.F,y=contrib[u][i].S.S;
		if (c>0) ans[id]+= cnt[c]*y-sum[c]+dist;
		else c=-c, ans[id]-= 2*(cnt[c]*y-sum[c]+dist);
	}
	for (int i=0;i<G[u].size();++i){
		int v=G[u][i].F,id=G[u][i].S;
		if (v!=pa){
			++cnt[col[id]];sum[col[id]]+=len[id];dist+=len[id];
			calc(v,u);
			--cnt[col[id]];sum[col[id]]-=len[id];dist-=len[id];
		}
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin>>n>>Q;
	REP(i,n-1){
		int x,y,c,d;cin>>x>>y>>col[i]>>len[i];
		G[x].pb(mp(y,i));G[y].pb(mp(x,i));
	}
	dfs(1,0);
	for (int j=0;j<18;++j) REP(i,n) anc[i][j+1]=anc[anc[i][j]][j];
	REP(i,Q){
		int x,y,u,v;cin>>x>>y>>u>>v;
		int lca=LCA(u,v);
		contrib[lca].pb(mp(i,mp(-x,y)));
		contrib[u].pb(mp(i,mp(x,y)));contrib[v].pb(mp(i,mp(x,y)));
	}
	calc(1,0);
	REP(i,Q) cout<<ans[i]<<endl;
	return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<int,ll> pil;
typedef pair<ll,int> pli;
#define rep(i,n) for (int i=0;i<n;++i)
#define REP(i,n) for (int i=1;i<=n;++i)
#define all(x) x.begin(),x.end()
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define read(x) scanf("%d",&x)
int n,Q;
vector<pii> G[100005];
int col[100005],len[100005];
int dep[100005];
int anc[100005][20];
inline void dfs(int u,int pa=0){
	dep[u]=dep[pa]+1;
	anc[u][0]=pa;
	for (int i=0;i<G[u].size();++i){
		int v=G[u][i].F;
		if (v!=pa) dfs(v,u);
	}
}
inline int LCA(int x,int y){
	if (dep[x]<dep[y]) swap(x,y);
	int d=dep[x]-dep[y];
	rep(j,19) if (d&(1<<j)) x=anc[x][j];
	if (x==y) return x;
	for (int j=18;j>=0;--j) if (anc[x][j]!=anc[y][j]) x=anc[x][j],y=anc[y][j];
	return anc[x][0];
}
ll ans[100005];
vector<pair<int,pii> > contrib[100005];// F:query id    S: F:col(-:double minus) S: y
ll cnt[100005],sum[100005],dist;
inline void calc(int u,int pa){
	for (int i=0;i<contrib[u].size();++i){
		int id=contrib[u][i].F,c=contrib[u][i].S.F,y=contrib[u][i].S.S;
		if (c>0) ans[id]+= cnt[c]*y-sum[c]+dist;
		else c=-c, ans[id]-= 2*(cnt[c]*y-sum[c]+dist);
	}
	for (int i=0;i<G[u].size();++i){
		int v=G[u][i].F,id=G[u][i].S;
		if (v!=pa){
			++cnt[col[id]];sum[col[id]]+=len[id];dist+=len[id];
			calc(v,u);
			--cnt[col[id]];sum[col[id]]-=len[id];dist-=len[id];
		}
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin>>n>>Q;
	REP(i,n-1){
		int x,y,c,d;cin>>x>>y>>col[i]>>len[i];
		G[x].pb(mp(y,i));G[y].pb(mp(x,i));
	}
	dfs(1,0);
	for (int j=0;j<18;++j) REP(i,n) anc[i][j+1]=anc[anc[i][j]][j];
	REP(i,Q){
		int x,y,u,v;cin>>x>>y>>u>>v;
		int lca=LCA(u,v);
		contrib[lca].pb(mp(i,mp(-x,y)));
		contrib[u].pb(mp(i,mp(x,y)));contrib[v].pb(mp(i,mp(x,y)));
	}
	calc(1,0);
	REP(i,Q) cout<<ans[i]<<endl;
	return 0;
}
```