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 ll long long
#define pii pair<int,int>
#define mp make_pair
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define real udnasjkd
#define fi first
#define se second
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
int now[100005],now1[100005];
vector<pair<int,pii> > g[100005];
int dep[100005],asc[100005][20];
map<pii,int> cnt,sum;
vector<int> ask[100005];
int n,m;
pair<pii,pii> in[100005];
ll real[100005];
void dfs(int u,int lst){
    dep[u]=dep[lst]+1;
    asc[u][0]=lst;
    for(int i=1;i<20;i++) asc[u][i]=asc[asc[u][i-1]][i-1];
    for(int i=0;i<(int)g[u].size();i++){
        int v=g[u][i].fi;
        if(v!=lst) dfs(v,u);
    }
}
int lca(int u,int v){
    if(dep[u]<dep[v]) swap(u,v);
    int i=19;
    while(dep[u]>dep[v]){
        for(;i>=0 && dep[asc[u][i]]<dep[v];i--);
        u=asc[u][i];
    }
    i=19;
    while(u!=v){
        for(;i>=0 && asc[u][i]==asc[v][i];i--);
        if(i==-1) i++;
        u=asc[u][i];
        v=asc[v][i];
    }
    return u;
}
void dfs1(int ind,int pa,int c,int d){
    now[c]+=d;
    now1[c]++;
    real[ind]=real[pa]+d;
    for(int i=0;i<ask[ind].size();i++){
        int nc=ask[ind][i];
        cnt[mp(ind,nc)]=now1[nc];
        sum[mp(ind,nc)]=now[nc];
    }
    for(int i=0;i<(int)g[ind].size();i++){
        int ha=g[ind][i].fi;
        if(ha!=pa)
            dfs1(ha,ind,g[ind][i].se.fi,g[ind][i].se.se);
    }
    now[c]-=d;
    now1[c]--;
}
int main(){
    scanf("%d %d",&n,&m);
    for(int i=1;i<n;i++){
        int a,b,c,d;
        scanf("%d %d %d %d",&a,&b,&c,&d);
        g[a].pb(mp(b,mp(c,d)));
        g[b].pb(mp(a,mp(c,d)));
    }
    dfs(1,0);
    for(int i=1;i<=m;i++){
        int a,b,c,d;
        scanf("%d %d %d %d",&a,&b,&c,&d);
        int ha=lca(c,d);
        ask[c].pb(a);
        ask[d].pb(a);
        ask[ha].pb(a);
        in[i]=mp(mp(a,b),mp(c,d));
    }
    dfs1(1,0,0,0);
    ll ans=0;
    for(int i=1;i<=m;i++){
        int x=cnt[mp(in[i].se.fi,in[i].fi.fi)]*in[i].fi.se-sum[mp(in[i].se.fi,in[i].fi.fi)]+real[in[i].se.fi];
        int y=cnt[mp(in[i].se.se,in[i].fi.fi)]*in[i].fi.se-sum[mp(in[i].se.se,in[i].fi.fi)]+real[in[i].se.se];
        int ha=lca(in[i].se.fi,in[i].se.se);
        int z=cnt[mp(ha,in[i].fi.fi)]*in[i].fi.se-sum[mp(ha,in[i].fi.fi)]+real[ha];
        //cout<<x<<" "<<y<<" "<<z<<endl;
        cout<<x+y-2*z<<endl;
        ans+=x+y-2*z;
    }
    return 0;
}