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 ll long long
#define mp make_pair
#define pb push_back

const int N = 1e5 + 5;
int n, q;
vector <int> g[N], col[N], cost[N];
int dep[N], siz[N], son[N], top[N], id[N], cnt;
int dfn[N], on[N], tme, fa[N], fae[N];
vector <int> occ[N];
vector <int> sum[N];
int sum_cost[N];

void dfs0(int u, int p, int pi) {
	if (~p) dep[u] = dep[p] + 1;
	siz[u] = 1; son[u] = -1; fa[u] = p; fae[u] = pi;
	for (int i = 0; i < g[u].size(); ++ i) {
		int v = g[u][i];
		if (v != p) {
			dfs0(v, u, i);
			siz[u] += siz[v];
			if (!~son[u] || siz[v] > siz[son[u]]) son[u] = v;
		}
	}
}

void dfs1(int u, int p) {
	id[u] = cnt;
	dfn[u] = tme; on[tme ++] = u;
	if (!~top[cnt]) top[cnt] = u;
	if (~son[u]) dfs1(son[u], u);
	for (int i = 0; i < g[u].size(); ++ i) {
		int v = g[u][i];
		if (v != p && v != son[u]) {
			++ cnt;
			dfs1(v, u);
		}
	}
}

int main() {
	scanf("%d %d", &n, &q);
	for (int i = 1; i < n; ++ i) {
		int u, v, w, c;
		scanf("%d %d %d %d", &u, &v, &c, &w);
		-- u; -- v; -- c;
		g[u].pb(v); col[u].pb(c); cost[u].pb(w);
		g[v].pb(u); col[v].pb(c); cost[v].pb(w);
	}
	memset(top, -1, sizeof(top));
	dfs0(0, -1, -1); dfs1(0, -1);
	for (int i = 0; i < n; ++ i) {
		int u = on[i];
		if (!u) continue;
		occ[col[fa[u]][fae[u]]].pb(i);
		int lst = sum[col[fa[u]][fae[u]]].empty() ? 0 : sum[col[fa[u]][fae[u]]].back();
		sum[col[fa[u]][fae[u]]].pb(lst + cost[fa[u]][fae[u]]);
		sum_cost[i] = (i ? sum_cost[i - 1] : 0) + cost[fa[u]][fae[u]];
	}
	while (q --) {
		int c, w, u, v;
		scanf("%d %d %d %d", &c, &w, &u, &v);
		-- u; -- v; -- c;
		int dis = 0, cnt = 0, res = 0;
		while (id[u] != id[v]) {
			if (dep[top[id[u]]] < dep[top[id[v]]]) swap(u, v);
			dis += sum_cost[dfn[u]] - (dfn[top[id[u]]] ? sum_cost[dfn[top[id[u]]] - 1] : 0);
			int L = lower_bound(occ[c].begin(), occ[c].end(), dfn[top[id[u]]]) - occ[c].begin();
			int R = upper_bound(occ[c].begin(), occ[c].end(), dfn[u]) - occ[c].begin() - 1;
			cnt += R - L + 1;
			if (L <= R) res += sum[c][R] - (L ? sum[c][L - 1] : 0);
			u = fa[top[id[u]]];
		}
		if (dep[u] > dep[v]) swap(u, v);
		dis += sum_cost[dfn[v]] - sum_cost[dfn[u]];
		int L = upper_bound(occ[c].begin(), occ[c].end(), dfn[u]) - occ[c].begin();
		int R = upper_bound(occ[c].begin(), occ[c].end(), dfn[v]) - occ[c].begin() - 1;
		cnt += R - L + 1;
		if (L <= R) res += sum[c][R] - (L ? sum[c][L - 1] : 0);
		printf("%d\n", dis - res + cnt * w);
	}
	return 0;
}