#include<bits/stdc++.h>
using namespace std;
int n,m,h[101010],f[101010][22],dt[101010],c[101010],cc[101010],ans[101010];
vector<pair<int,pair<int,int> > >v[101010],q[101010];
void gt(int x)
{
	h[x]=h[f[x][0]]+1;
	for(int i=0;i<v[x].size();i++)
	{
		if(v[x][i].first!=f[x][0])
		{
			f[v[x][i].first][0]=x;
			dt[v[x][i].first]=dt[x]+v[x][i].second.second;
			gt(v[x][i].first);
		}
	}
}
int hb(int x,int y)
{
	if(h[x]>h[y])
	{
		swap(x,y);
	}
	int ch=h[y]-h[x];
	for(int i=0;i<=20;i++)
	{
		if(ch&(1<<i))
		{
			y=f[y][i];
		}
	}
	if(x==y)
	{
		return x;
	}
	for(int i=20;i>=0;i--)
	{
		if(f[x][i]!=f[y][i])
		{
			x=f[x][i];
			y=f[y][i];
		}
	}
	return f[x][0];
}
void dfs(int x)
{
	for(int i=0;i<q[x].size();i++)
	{
		if(q[x][i].first>0)
		{
			ans[q[x][i].first]+=dt[x]+c[q[x][i].second.first]*q[x][i].second.second-cc[q[x][i].second.first];
		}
		else
		{
			ans[-q[x][i].first]-=2*(dt[x]+c[q[x][i].second.first]*q[x][i].second.second-cc[q[x][i].second.first]);
		}
	}
	for(int i=0;i<v[x].size();i++)
	{
		if(v[x][i].first==f[x][0])
		{
			continue;
		}
		c[v[x][i].second.first]++;
		cc[v[x][i].second.first]+=v[x][i].second.second;
		dfs(v[x][i].first);
		c[v[x][i].second.first]--;
		cc[v[x][i].second.first]-=v[x][i].second.second;
	}
}
int main()
{
	cin>>n>>m;
	for(int i=0;i<n-1;i++)
	{
		int a,b,c,d;
		cin>>a>>b>>c>>d;
		v[a].push_back(make_pair(b,make_pair(c,d)));
		v[b].push_back(make_pair(a,make_pair(c,d)));
	}
	gt(1);
	for(int i=1;i<=20;i++)
	{
		for(int j=1;j<=n;j++)
		{
			f[j][i]=f[f[j][i-1]][i-1];
		}
	}
	for(int i=1;i<=m;i++)
	{
		int c,d,x,y;
		cin>>c>>d>>x>>y;
		q[x].push_back(make_pair(i,make_pair(c,d)));
		q[y].push_back(make_pair(i,make_pair(c,d)));
		q[hb(x,y)].push_back(make_pair(-i,make_pair(c,d)));
	}
	dfs(1);
	for(int i=1;i<=m;i++)
	{
		cout<<ans[i]<<endl;
	}
	return 0;
}
### Prompt

Your task is to create 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;
int n,m,h[101010],f[101010][22],dt[101010],c[101010],cc[101010],ans[101010];
vector<pair<int,pair<int,int> > >v[101010],q[101010];
void gt(int x)
{
	h[x]=h[f[x][0]]+1;
	for(int i=0;i<v[x].size();i++)
	{
		if(v[x][i].first!=f[x][0])
		{
			f[v[x][i].first][0]=x;
			dt[v[x][i].first]=dt[x]+v[x][i].second.second;
			gt(v[x][i].first);
		}
	}
}
int hb(int x,int y)
{
	if(h[x]>h[y])
	{
		swap(x,y);
	}
	int ch=h[y]-h[x];
	for(int i=0;i<=20;i++)
	{
		if(ch&(1<<i))
		{
			y=f[y][i];
		}
	}
	if(x==y)
	{
		return x;
	}
	for(int i=20;i>=0;i--)
	{
		if(f[x][i]!=f[y][i])
		{
			x=f[x][i];
			y=f[y][i];
		}
	}
	return f[x][0];
}
void dfs(int x)
{
	for(int i=0;i<q[x].size();i++)
	{
		if(q[x][i].first>0)
		{
			ans[q[x][i].first]+=dt[x]+c[q[x][i].second.first]*q[x][i].second.second-cc[q[x][i].second.first];
		}
		else
		{
			ans[-q[x][i].first]-=2*(dt[x]+c[q[x][i].second.first]*q[x][i].second.second-cc[q[x][i].second.first]);
		}
	}
	for(int i=0;i<v[x].size();i++)
	{
		if(v[x][i].first==f[x][0])
		{
			continue;
		}
		c[v[x][i].second.first]++;
		cc[v[x][i].second.first]+=v[x][i].second.second;
		dfs(v[x][i].first);
		c[v[x][i].second.first]--;
		cc[v[x][i].second.first]-=v[x][i].second.second;
	}
}
int main()
{
	cin>>n>>m;
	for(int i=0;i<n-1;i++)
	{
		int a,b,c,d;
		cin>>a>>b>>c>>d;
		v[a].push_back(make_pair(b,make_pair(c,d)));
		v[b].push_back(make_pair(a,make_pair(c,d)));
	}
	gt(1);
	for(int i=1;i<=20;i++)
	{
		for(int j=1;j<=n;j++)
		{
			f[j][i]=f[f[j][i-1]][i-1];
		}
	}
	for(int i=1;i<=m;i++)
	{
		int c,d,x,y;
		cin>>c>>d>>x>>y;
		q[x].push_back(make_pair(i,make_pair(c,d)));
		q[y].push_back(make_pair(i,make_pair(c,d)));
		q[hb(x,y)].push_back(make_pair(-i,make_pair(c,d)));
	}
	dfs(1);
	for(int i=1;i<=m;i++)
	{
		cout<<ans[i]<<endl;
	}
	return 0;
}
```