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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define SZ(a) (int)(a.size())
#define PB push_back
#define LL long long
#define X first
#define Y second

const int MAX = 100005;
const int MAGIC = 350;
vector<vector<pair<int , pair<int , int> > > > g(MAX);
int timer;
int cnt[MAX];
int tin[MAX];
int tout[MAX];
int D[MAX];
int dp[MAX][18];
pair<int , int> d[400][MAX];
int idx[MAX];
vector<vector<pair<pair<int , int> , int> > > edges(MAX);
bool isIn(int u , int v)
{
	return tin[u] >= tin[v] && tout[u] <= tout[v];
}
int lca(int u , int v)
{
	if(isIn(u , v))
		return v;
	if(isIn(v , u))
		return u;
	for(int i = 17; i >= 0; i--)
	{
		if(!isIn(v , dp[u][i]))
		{
			u = dp[u][i];
		}
	}
	return dp[u][0];
}
void dfs(int u, int p = -1)
{
	tin[u] = timer++;
	for(auto v : g[u])
	{
		if(v.X == p)
			continue;
		D[v.X] = D[u] + v.Y.X;
		dp[v.X][0] = u;
		for(int i = 1; i <= 17; i++)
			dp[v.X][i] = dp[dp[v.X][i - 1]][i - 1];
		dfs(v.X , u);
	}
	tout[u] = timer++;
}
void dfs2(int u , int i, int idx, int par = -1)
{
	for(auto v : g[u])
	{
		if(v.X == par)
			continue;
		d[idx][v.X] = {d[idx][u].X + (v.Y.Y == i) , d[idx][u].Y + (v.Y.Y == i) * v.Y.X};
		dfs2(v.X , i , idx , u);
	}
}
int main()
{
	int n , q;
	cin >> n >> q;
	for(int i = 1; i < n; i++)
	{
		int u , v , c , d;
		cin >> u >> v >> c >> d;
		u--;v--;
		g[u].PB({v , {d , c}});
		g[v].PB({u , {d , c}});
		edges[c].PB({{u , v} , d});
	}
	dfs(0);
	memset(idx , -1 , sizeof idx);
	int ptr = 0;
	for(int i = 1; i < n; i++)
		if(SZ(edges[i]) >= MAGIC)
		{
			idx[i] = ptr++;
			dfs2(0 , i , idx[i]);
		}
	while(q--)
	{
		int x , y , u , v;
		cin >> x >> y >> u >> v;
		u--;v--;
		int LCA = lca(u , v);
		if(idx[x] == -1)//small
		{
			int ans = D[u] + D[v] - 2 * D[LCA];
			for(auto e : edges[x])
			{
				if(isIn(e.X.X , LCA) && isIn(e.X.Y , LCA) && (isIn(u , e.X.X) && isIn(u , e.X.Y) || isIn(v , e.X.X) && isIn(v , e.X.Y)))
				{
					ans -= e.Y;
					ans += y;
				}
			}
			cout << ans << endl;
		}
		else
		{
			int ans = D[u] + D[v] - 2 * D[LCA];
			//cerr << ans << endl;
			ans -= d[idx[x]][u].Y + d[idx[x]][v].Y - 2 * d[idx[x]][LCA].Y;
			//cerr << ans << endl;
			ans += y * (d[idx[x]][u].X + d[idx[x]][v].X - 2 * d[idx[x]][LCA].X);
			cout << ans << endl;
		}
	}
    return 0;
}