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;

const int N = 100005;
const int L = 18;

vector < pair < int, pair < int, int > > > g[N];
vector < pair < int, pair < int, pair < int, int > > > > queries[N];
int ant[N][L], nv[N], ans[N], dist[N];
vector < int > rememberList[N];
map < int, int > remember[2][N];
map < int, int > now[2];

void build(int u, int p)
{
	ant[u][0] = p;
	for (int i=1; i<L; i++){
		ant[u][i] = ant[ant[u][i-1]][i-1];
	}
	for (pair < int, pair < int, int > > v : g[u])
	{
		if(v.first == p) continue ;
		nv[v.first] = nv[u]+1;
		dist[v.first] = dist[u] + v.second.second;
		build(v.first, u);
	}
}

int lca(int u, int v)
{
	if(nv[u] < nv[v]) swap(u, v);
	for (int i=L-1; i>=0; i--)
	{
		if(nv[ant[u][i]] >= nv[v]){
			u = ant[u][i];
		}
	}
	if(u == v) return u;
	for (int i=L-1; i>=0; i--)
	{
		if(ant[u][i] != ant[v][i])
		{
			u = ant[u][i], v = ant[v][i];
		}
	}
	return ant[v][0];
}

void solve(int u, int p)
{
	for (int x : rememberList[u])
	{
		remember[0][u][x] = now[0][x];
		remember[1][u][x] = now[1][x];
	}

	for (auto x : queries[u]){
		int LCA = x.first;
		int id = x.second.second.second;
		ans[id] -= (now[1][x.second.first] - remember[1][LCA][x.second.first]);
		ans[id] += ((now[0][x.second.first] - remember[0][LCA][x.second.first]) * x.second.second.first);
		// cout << id << ' ' << u << ' ' << remember[1][LCA][x.second.first] << ' ' << 
	}

	for (pair < int, 	pair < int, int > > v : g[u])
	{
		if(v.first == p) continue ;
		now[0][v.second.first]++;
		now[1][v.second.first] += v.second.second;
		solve(v.first, u);
		now[0][v.second.first]--;
		now[1][v.second.first] -= v.second.second;
	}
}

int main()
{
	int n, q;
	scanf("%d %d", &n, &q);
	for (int i=0; i<n-1; i++){
		int u, v, c, d;
		scanf("%d %d %d %d", &u, &v, &c, &d);
		g[u].push_back({v, {c, d}});
		g[v].push_back({u, {c, d}});
	}
	nv[1] = 0;
	build(1, 1);
	for (int i=0; i<q; i++){
		int x, y, u, v;
		scanf("%d %d %d %d", &x, &y, &u, &v);
		int LCA = lca(u, v);
		rememberList[LCA].push_back(x);
		queries[u].push_back({LCA, {x, {y, i}}});
		queries[v].push_back({LCA, {x, {y, i}}});
		ans[i] = dist[u] + dist[v] - 2*dist[LCA];
	}
	solve(1, 1);
	for (int i=0; i<q; i++)
	{
		printf("%d\n", ans[i]);
	}
}