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;

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
	for (; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

const int N = 1e5 + 5;
struct node{ int v, c, d; };
vector<node> G[N], Q[N];
int dep[N], par[N][20], dis[N];
int cnt[N], sum[N], ans[N];

void dfs(int u, int fa)
{
	dep[u] = dep[fa] + 1;
	par[u][0] = fa;
	for (int i = 1; i <= 18; ++i)
		par[u][i] = par[par[u][i - 1]][i - 1];
	
	for (auto i : G[u]) if (i.v != fa)
		dis[i.v] = dis[u] + i.d, dfs(i.v, u);
}

int LCA(int u, int v)
{
	if (dep[u] > dep[v]) swap(u, v);
	for (int i = 18; i >= 0; --i)
		if (dep[par[v][i]] >= dep[u])
			v = par[v][i];
	if (u == v) return u;
	
	for (int i = 18; i >= 0; --i)
		if (par[u][i] != par[v][i])
			u = par[u][i], v = par[v][i];
	return par[u][0];
}

void solve(int u, int fa)
{
	for (auto i : Q[u])
	{
		if (i.v > 0) ans[i.v] += dis[u] - sum[i.c] + cnt[i.c] * i.d;
		else ans[-i.v] -= 2 * (dis[u] - sum[i.c] + cnt[i.c] * i.d);
	}
	
	for (auto i : G[u]) if (i.v != fa)
	{
		cnt[i.c]++; sum[i.c] += i.d;
		solve(i.v, u);
		cnt[i.c]--; sum[i.c] -= i.d;
	}
}

int main()
{
	int n = read();
	int q = read();
	for (int i = 1; i < n; ++i)
	{
		int u = read(), v = read();
		int c = read(), d = read();
		G[u].push_back((node){v, c, d});
		G[v].push_back((node){u, c, d});
	}
	dfs(1, 0);
	
	for (int i = 1; i <= q; ++i)
	{
		int c = read(), d = read();
		int u = read(), v = read();
		Q[u].push_back((node){i, c, d});
		Q[v].push_back((node){i, c, d});
		Q[LCA(u, v)].push_back((node){-i, c, d});
	}
	solve(1, 0);
	
	for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
	
	return 0;
}