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;
typedef tuple<int,int,int>T;
typedef pair<int,int>P;
const int N=100000,M=262144,inf=2000000000;
vector<vector<T>>e(N);
vector<vector<P>>tour_d(N);
vector<int>tour_v;
vector<int>segt(M*2,inf);
vector<vector<int>>sum(N-1,vector<int>(1));
vector<vector<int>>num(N-1,vector<int>(1));
vector<vector<int>>p(N-1,vector<int>(1));
int dist[N],newv[N],t[N];
int cnt=0,vis=0;
void dfs(int v,int before){
    int nowv=cnt;
    newv[v]=cnt;
    if(t[cnt]==0)t[cnt]=vis;
    cnt++,vis++;
    tour_v.push_back(nowv);
    for(int i=0;i<e[v].size();i++){
        int u=get<0>(e[v][i]),col=get<1>(e[v][i]),len=get<2>(e[v][i]);
        if(u!=before){
            dist[cnt]=dist[nowv]+len;
            tour_d[col].push_back({len,1});
            p[col].push_back(vis);
            dfs(u,v);
            tour_v.push_back(nowv);
            tour_d[col].push_back({-len,-1});
            p[col].push_back(vis);
            vis++;
        }
    }
}
int segmin(int l,int r,int n,int nl,int nr){
    int mid=(nl+nr)/2;
    if(r<=nl || nr<=l)return inf;
    else if(l<=nl && nr<=r)return segt[n];
    else return min(segmin(l,r,n*2,nl,mid),segmin(l,r,n*2+1,mid,nr));
}
int dis(int v,int x,int y){
    v=upper_bound(p[x].begin(),p[x].end(),v)-p[x].begin()-1;
    return num[x][v]*y-sum[x][v];
}
int main(){
    int n,q;
    cin>>n>>q;
    for(int i=0;i<n-1;i++){
        int a,b,c,d;
        cin>>a>>b>>c>>d;
        e[a-1].emplace_back(b-1,c-1,d);
        e[b-1].emplace_back(a-1,c-1,d);
    }
    dfs(0,-1);
    for(int i=0;i<tour_v.size();i++)segt[M+i]=tour_v[i];
    for(int i=M-1;i;i--)segt[i]=min(segt[i*2],segt[i*2+1]);
    for(int i=0;i<n-1;i++){
        for(int j=0;j<tour_d[i].size();j++){
            sum[i].push_back(sum[i][j]+tour_d[i][j].first);
            num[i].push_back(num[i][j]+tour_d[i][j].second);
        }
    }
    int ans[q];
    for(int i=0;i<q;i++){
        int u,v,x,y;
        cin>>x>>y>>u>>v;
        u=newv[u-1],v=newv[v-1],x--;
        int lca=segmin(min(t[u],t[v]),max(t[u],t[v])+1,1,0,M);
        ans[i]=dis(t[u],x,y)+dis(t[v],x,y)-2*dis(t[lca],x,y)
              +dist[u]+dist[v]-2*dist[lca];
    }
    for(int i=0;i<q;i++)cout<<ans[i]<<endl;
    return 0;
}