Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

130
200
60

The graph in this input is as follows:

Here the edges of Color 1 are shown as solid red lines, the edge of Color 2 is shown as a bold green line, and the edge of Color 4 is shown as a blue dashed line.

Query 1: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 4 is 100 + 30 = 130.

Query 2: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 5 is 100 + 100 = 200.

Query 3: Assuming that the length of every edge whose color is 3 is changed to 1000 (there is no such edge), the distance between Vertex 3 and Vertex 4 is 20 + 10 + 30 = 60. Note that the edges of Color 1 now have their original lengths.

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<string>
#include<iomanip>
#include<cmath>
#include<vector>
#include<algorithm>
#include<queue>
#include<cassert>

using namespace std;

#define int long long
#define rep(i,n) for(int i = 0; i < (n); i++)
#define endl "\n"

const long long INF = (long long)1e18;
const long long MOD = (long long)1e9 + 7; 

string yn(bool f){return f?"Yes":"No";}
string YN(bool f){return f?"YES":"NO";}

#define MAX  210000


typedef long long ll;
typedef pair<ll,ll> P;

struct edge{ll to, cost, col;};

vector<edge> G[MAX];
ll d[MAX], V;

int dijkstra(int x, int y, ll s, int g){
	priority_queue<P,vector<P>,greater<P>> q;
	fill(d,d+V, INF);
	assert(V < MAX);
	
	d[s] = 0;
	q.push(P(0,s));
	assert(s < MAX);
	while(!q.empty()){ 
		P p = q.top();q.pop();
		ll v = p.second;assert(v< MAX);
		// cout<<"v = "<<v<<" g = "<<g<<endl;
		if(v == g) return d[g];assert(g < MAX);
		
		if(d[v] < p.first) continue;
		
		for(edge e : G[v]){
			int cost = e.cost;
			// cout<<e.to<<endl;
			if(e.col == x) cost = y;
			
			// cout<<"cost "<<cost<<endl;
			assert(e.to < MAX);
			if(d[e.to] > d[v] + cost){
				d[e.to] = d[v] + cost;
				q.push(P(d[e.to],e.to));
			}
		}
	}
	assert(g < MAX);
	return d[g];
}
signed main(){
	// cin.tie(0);
	// ios::sync_with_stdio(false);
	// cout<<fixed<<setprecision(10);
	
	int N, Q;
	int a, b, c, d;
	int y, x, u, v;
	
	
	
	cin>>N>>Q;
	
	V = N+10;
	
	for(int i = 0; i < N-1; i++){
		cin>>a>>b>>c>>d;
		edge e;
		e.to = b;
		e.col = c;
		e.cost = d;
		assert(a < MAX);
		G[a].push_back(e);
		
		e.to = a;
		assert(b < MAX);
		G[b].push_back(e);
	}
	
	for(int i = 0; i < Q; i++){
		cin>>x>>y>>u>>v;
		
		cout<<dijkstra(x,y,u,v)<<endl;
	}
	
	return 0;
}