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 pb push_back
const int maxn=3e5+5;
const ll p=1000000007;
vector<pair<ll,pair<ll,ll> > >G[maxn];
vector<pair<ll,pair<ll,pair<ll,ll> > > >Q[maxn];

int dep[maxn],f[maxn][25];
void DFS(int x, int fa) {
    dep[x] = dep[fa] + 1;
    f[x][0] = fa;
    for (pair<ll,pair<ll,ll> > p:G[x])
	    if(!dep[p.first]) DFS(p.first, x);
}
 
int n;

void beizeng() {
    DFS(1, 0);
    for(int j = 1; j <= 24; j++)
        for(int i = 1; i <= 2*n; i++)
            f[i][j] = f[f[i][j - 1]][j - 1];
}
 
int LCA(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    for(int i = 24; ~i; i--) if(dep[f[x][i]] >= dep[y]) x = f[x][i];
    if(x == y) return x;
    for(int i = 24; ~i; i--) if(f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
    return f[x][0];
}
 
 

ll ans[maxn];
ll cnt[maxn],sum[maxn];

void dfs(int u,int fa,ll len){
	
	for (pair<ll,pair<ll,pair<ll,ll> > >p:Q[u]){
		ans[p.second.first]+=p.first*(len-sum[p.second.second.first]+cnt[p.second.second.first]*p.second.second.second);
	}
	
	for (pair<ll,pair<ll,ll> > p:G[u]){
		if (p.first==fa) continue;
		cnt[p.second.first]++;
		sum[p.second.first]+=p.second.second;
		dfs(p.first,u,len+p.second.second);
		cnt[p.second.first]--;
		sum[p.second.first]-=p.second.second;
	}
	
	
}

int main(){
	ios::sync_with_stdio(false);

	int q;
	cin>>n>>q;
	
	for (int i=0;i<n-1;i++){
		ll u,v,c,d;
		cin>>u>>v>>c>>d;
		G[u].pb({v,{c,d}});
		G[v].pb({u,{c,d}});
	}
	
	beizeng();
	
	for (int i=1;i<=q;i++){
		int x,y,u,v;
		cin>>x>>y>>u>>v;
		int lca=LCA(u,v);
		Q[lca].pb({-2,{i,{x,y}}});
		Q[u].pb({1,{i,{x,y}}});
		Q[v].pb({1,{i,{x,y}}});
	}
	
	dfs(1,0,0);
	
	for (int i=1;i<=q;i++){
		cout<<ans[i]<<endl;
	}
}