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 <stdio.h>
#include <vector>
using namespace std;

struct edge{
	int y, c, d;
};
vector<edge> G[100100], V[100100];

const int Z = 16;
int N, Q, C, S[100100], E[100100], D[100100], P[100100][Z+1];
int bit[720][100100], B, I[100100];

void go(int x, int l)
{
	S[x] = ++C;

	D[x] = D[l] + 1;
	P[x][0] = l;
	for (int i = 1; i <= Z; i++) P[x][i] = P[P[x][i - 1]][i - 1];

	for (auto &e : G[x]) if (e.y != l){
		V[e.c].push_back(e);
		go(e.y, x);
	}

	E[x] = C + 1;
}

int lca(int x, int y)
{
	if (D[x] > D[y]) swap(x, y);

	int d = D[y] - D[x];
	for (int i = 0; i <= Z; i++) if (d & (1 << i)) y = P[y][i];

	if (x != y){
		for (int i = Z; i >= 0; i--) if (P[x][i] != P[y][i]){
			x = P[x][i];
			y = P[y][i];
		}
		x = P[x][0];
	}
	return x;
}

void push(int *b, int x, int p)
{
	while (x <= N){
		b[x] += p;
		x += x & (-x);
	}
}

int pop(int *b, int x)
{
	int r = 0;
	while (x){
		r += b[x];
		x -= x & (-x);
	}
	return r;
}

int main()
{
	scanf ("%d %d", &N, &Q);
	for (int i = 1; i < N; i++){
		int x, y, c, d; scanf ("%d %d %d %d", &x, &y, &c, &d);
		G[x].push_back({ y,c,d });
		G[y].push_back({ x,c,d });
	}

	go(1, 0);
	
	for (int i = 1; i < N; i++){
		for (auto &e : V[i]){
			push(bit[0], S[e.y], e.d);
			push(bit[0], E[e.y], -e.d);
		}
		if (V[i].size() >= 300){
			I[i] = ++B;
			for (auto &e : V[i]){
				push(bit[B], S[e.y], -e.d);
				push(bit[B], E[e.y], e.d);
				push(bit[360+B], S[e.y], 1);
				push(bit[360+B], E[e.y], -1);
			}
		}
	}

	while (Q--){
		int c, d, x, y;
		scanf ("%d %d %d %d", &c, &d, &x, &y);
		int z = lca(x, y);

		if (V[c].size() < 300){
			for (auto &e : V[c]){
				push(bit[0], S[e.y], d - e.d);
				push(bit[0], E[e.y], e.d - d);
			}
			int r = pop(bit[0], S[x]) + pop(bit[0], S[y]) - 2 * pop(bit[0], S[z]);
			printf ("%d\n", r);
			for (auto &e : V[c]){
				push(bit[0], S[e.y], e.d - d);
				push(bit[0], E[e.y], d - e.d);
			}
		}
		else{
			int i = I[c];
			int r = pop(bit[0], S[x]) + pop(bit[0], S[y]) - 2 * pop(bit[0], S[z]);
			r += pop(bit[i], S[x]) + pop(bit[i], S[y]) - 2 * pop(bit[i], S[z]);
			i += 360;
			r += (pop(bit[i], S[x]) + pop(bit[i], S[y]) - 2 * pop(bit[i], S[z])) * d;
			printf ("%d\n", r);
		}
	}

	return 0;
}