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>
#define fi first
#define se second
#define mp make_pair
using namespace std;

typedef pair <int, int> pii;
typedef pair <int, pii> pip;
typedef pair <pii, pii> ppp;

vector <pip> from[100010];
vector <ppp> qq[100010];
int fa[100010][17], dep[100010], dis[100010];
int sum[100010], sss[100010];
int ans[100010];

void dfs(int x, int last) {
	fa[x][0] = last;
	for (int j = 1; j < 17; j++) {
		fa[x][j] = fa[fa[x][j - 1]][j - 1];
	}
	for (int i = 0; i < from[x].size(); i++) {
		int v = from[x][i].fi, w = from[x][i].se.se;
		if (v == last) continue;
		dep[v] = dep[x] + 1, dis[v] = dis[x] + w;
		dfs(v, x);
	}
}

int LCA(int x, int y) {
	if (dep[x] < dep[y]) swap(x, y);
	int tmp = dep[x] - dep[y], cnt = 0;
	while (tmp) {
		if (tmp & 1) x = fa[x][cnt]; 
		tmp >>= 1, cnt++;
	}
	if (x == y) return x;
	for (int i = 16; i >= 0; i--) {
		if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
	}
	return fa[x][0];
}

void get_ans(int x, int last) {
	for (int i = 0; i < qq[x].size(); i++) {
		int id = qq[x][i].fi.fi, c = qq[x][i].fi.se, t1 = qq[x][i].se.fi, t2 = qq[x][i].se.se;
		ans[id] += t1 * sum[c] * t2 - t1 * sss[c];
	}
	for (int i = 0; i < from[x].size(); i++) {
		int v = from[x][i].fi, c = from[x][i].se.fi, w = from[x][i].se.se;
		if (v == last) continue;
		sum[c]++, sss[c] += w;
		get_ans(v, x);
		sum[c]--, sss[c] -= w;
	}
}

int main () {
	int n, q; 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);
		from[x].push_back(mp(y, mp(c, d)));
		from[y].push_back(mp(x, mp(c, d)));
	}
	dfs(1, 1);
	for (int i = 0; i < q; i++) {
		int x, y, c, d; scanf("%d%d%d%d", &c, &d, &x, &y);
		int lca = LCA(x, y);
		ans[i] = dis[x] + dis[y] - 2 * dis[lca];
		qq[x].push_back(mp(mp(i, c), mp(1, d)));
		qq[y].push_back(mp(mp(i, c), mp(1, d)));
		qq[lca].push_back(mp(mp(i, c), mp(-2, d)));
	}
	get_ans(1, 1);
	for (int i = 0; i < q; i++) {
		printf("%d\n", ans[i]);
	}
	return 0;
}