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 rep(i,n) for(int i=0;(i)<(n);++(i))
using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

const int limit = 100010;
using edge = struct{int to, color, cost;};
vector<edge> graph[limit];

int x[limit], y[limit], uv[limit][3];
int val[limit][3];

const int root = 0;
const int limit_log = 17;
int par[limit_log + 1][limit];
int depth[limit];

void dfs(int v, int p, int d){
	par[0][v] = p;
	depth[v] = d;
	for(auto &e: graph[v]) {
		if(e.to == p) continue;
		dfs(e.to, v, d + 1);
	}
}

using query = tuple<int, int, int, int>;
vector<query> que[limit];

int num[limit], total_cost[limit];

void dfs2(int v, int p, int dist){
	for(auto &q: que[v]){
		int color, cost, idx, pos;
		tie(color, cost, idx, pos) = q;
		val[idx][pos] = dist;
		val[idx][pos] -= total_cost[color];
		val[idx][pos] += num[color] * cost;
	}

	for(auto &e: graph[v]){
		if(e.to == p) continue;
		num[e.color]++;
		total_cost[e.color] += e.cost;
		dfs2(e.to, v, dist + e.cost);
		num[e.color]--;
		total_cost[e.color] -= e.cost;
	}
}

int main(void){
	int n, q;
	cin >> n >> q;

	rep(i, n - 1){
		int a, b, c, d;
		cin >> a >> b >> c >> d;
		a--, b--, c--;
		graph[a].push_back({b, c, d});
		graph[b].push_back({a, c, d});
	}

	dfs(root, -1, 0);
	
	rep(k, limit_log) rep(v, n){
		const int tmp = par[k][v];
		par[k + 1][v] = (tmp == -1) ? -1 : par[k][tmp];
	}

	rep(i, q){
		cin >> x[i] >> y[i] >> uv[i][0] >> uv[i][1];
		x[i]--, uv[i][0]--, uv[i][1]--;
		int cu = uv[i][0], cv = uv[i][1];

		if(depth[cu] - depth[cv] < 0){
			swap(cu, cv);
			swap(uv[i][0], uv[i][1]);
		}

		int diff = depth[cu] - depth[cv];

		rep(_k, limit_log + 1) {
			const int k = limit_log - _k;
			if(diff >> k &1){
				cu = par[k][cu];
				diff -= (1 << k);
			}
		}

		if (cu != cv) {
			rep(_k, limit_log + 1) {
				const int k = limit_log - _k;
				if(par[k][cu] != par[k][cv]){
					cu = par[k][cu];
					cv = par[k][cv];
				}
			}
		}

		uv[i][2] = (cu == cv) ? cu : par[0][cu];
		rep(j, 3) que[uv[i][j]].push_back(query(x[i], y[i], i, j));
	}

	dfs2(root, -1, 0);

	rep(i, q){
		const int ans = val[i][0] + val[i][1] - 2 * val[i][2];
		cout << ans << endl;
	}

	return 0;
}