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 debug //printf
using namespace std;

const int maxn = 1e5+10;

int n;

struct edge{ int a, b, c; };
struct query{ int i, v, x, y; };
vector<edge> g[maxn];
vector<query> q[maxn];
void add2(int a, int b, int c, int d) {
	g[a].push_back((edge){b, c, d});
	g[b].push_back((edge){a, c, d});
}

int f[maxn], v[maxn];
int find(int x) { return f[x]>0?f[x]=find(f[x]):x; }
void merge(int x, int y) {
	int fx = find(x), fy = find(y);
	if (fx == fy) return;
	v[fy] = v[fx];
	if (f[fx] > f[fy]) swap(fx, fy);
	f[fx] -= f[fx]==f[fy];
	f[fy] = fx;
}

int dis[maxn], col[maxn], cdis[maxn];
int ans[maxn], lca[maxn];
void dfs(int u, int fa) {
	for (int i = 0; i < g[u].size(); ++i) {
		int v = g[u][i].a;
		if (v == fa) continue;
		dis[v] = dis[u] + g[u][i].c;
		col[g[u][i].b]++;
		cdis[g[u][i].b] += g[u][i].c;
		dfs(v, u);
		merge(u, v);
		col[g[u][i].b]--;
		cdis[g[u][i].b] -= g[u][i].c;
	}
	for (int i = 0; i < q[u].size(); ++i) {
		query &qq = q[u][i];
		debug(" %d %d %d %d: %d\n", qq.i, qq.v, qq.x, qq.y, ans[qq.i]);
		debug("%d: %d %d %d %d\n", qq.i, u, dis[u], cdis[qq.x], col[qq.x]*qq.y);
		if (qq.v) {
			ans[qq.i] += dis[u]-cdis[qq.x]+col[qq.x]*qq.y;
			if (!lca[qq.i] && dis[qq.v]) {
				lca[qq.i] = v[find(qq.v)];
				debug("lca: %d %d: %d\n", u, qq.v, lca);
				q[lca[qq.i]].push_back((query){qq.i, 0, qq.x, qq.y});
			}
		}
		else {
			ans[qq.i] -= 2*(dis[u]-cdis[qq.x]+col[qq.x]*qq.y);
		}
	}
}

int main() {
	int qq;
	scanf("%d%d", &n, &qq);
	int a, b, c, d;
	for (int i = 0; i < n-1; ++i) {
		scanf("%d%d%d%d", &a, &b, &c, &d);
		add2(a, b, c, d);
	}
	for (int i = 0; i < qq; ++i) {
		scanf("%d%d%d%d", &a, &b, &c, &d);
		q[c].push_back((query){i, d, a, b});
		q[d].push_back((query){i, c, a, b});
	}
	memset(f, -1, sizeof(f));
	for (int i = 1; i <= n; ++i) v[i] = i;
	dis[1] = 1;
	dfs(1, 0);
	for (int i = 0; i < qq; ++i) {
		printf("%d\n", ans[i]);
	}
}