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 <vector>
#include <iostream>
#include <queue>
using namespace std;

#define Pb push_back
/*
 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
 */
const int maxn =3e5+10;
struct Edge{
    int to,corlor,w;
};
vector<Edge> G[maxn];

int depth[maxn],vs[maxn],ID[maxn],dis[maxn];
int n;
int f[maxn][20];
void dfs(int node,int fa,int d,int &k){
    ID[node] = k;
    vs[k] = node;
    depth[k++] = d;
    // dis[node] = distance;
    for(int i = 0;i < G[node].size(); ++i){
        Edge &t = G[node][i];
        if(t.to == fa) continue;
        dis[t.to] = dis[node]+t.w;
        dfs(t.to,node,d+1,k);
        vs[k] = node;
        depth[k++] = d;
    }
}


void init_rmq(int n){
    for(int i = 1;i <= n ; ++i) f[i][0] = i;
    for(int j = 1;(1<<j) <= n; ++j){
        for(int i = 1;i + (1<<j)-1 <= n; ++i){
            if(depth[f[i][j-1]]< depth[f[i+(1<<(j-1))][j-1]])
                f[i][j] = f[i][j-1];
            else
                f[i][j] = f[i+(1<<(j-1))][j-1];
            
        }
    }
}
int rmq_query(int l,int r){
    int x = 0;
    while((1<<(x+1)) <= r-l+1) x++;
    if(depth[f[l][x]] < depth[f[r-(1<<x)+1][x]])
        return f[l][x];
    else
        return f[r-(1<<x)+1][x];
}
void init_lca(){
    int k = 1;
    dfs(1,-1,0,k);
    init_rmq(2*n-1);
}
int lca(int u,int v){
    // cout<<rmq_query(ID[u],ID[v])<<endl;
    return vs[rmq_query(min(ID[u],ID[v]),max(ID[u],ID[v]))];
}
struct que{
    int corlor,w,coe,id;
};
vector<que> Query[maxn];
int ans[maxn],num[maxn],value[maxn];// num[i] 个数，value 值
void dfs(int u,int fa){
    for(auto qu:Query[u]){
        // cout<<dis[u]<<endl;
        ans[qu.id] +=(dis[u]+num[qu.corlor]*qu.w-value[qu.corlor])*qu.coe;
    }
    for(auto p: G[u]){
        int to = p.to;
        if(to == fa) continue;
        num[p.corlor]++;
        value[p.corlor] += p.w;
        dfs(to,u);
        num[p.corlor]--;
        value[p.corlor] -= p.w;
    }
}
int main(void)
{
    //cout<<"abc177F Colorful Tree\n";
    int q;
    cin>>n>>q;
    for(int i = 1;i < n;++i){
        int u,v,c,w;
        scanf("%d%d%d%d",&u,&v,&c,&w);
        G[u].Pb(Edge{v,c,w});
        G[v].Pb(Edge{u,c,w});
    }
    init_lca();
    for(int i = 1;i <= q; ++i){
        int x,y,u,v;
        scanf("%d%d%d%d",&x,&y,&u,&v);
        int fa = lca(u,v);
        
        Query[u].Pb(que{x,y,1,i});
        Query[v].Pb(que{x,y,1,i});
        Query[fa].Pb(que{x,y,-2,i});
    }
    dfs(1,-1);
    for(int i  = 1;i <= q; ++i){
        printf("%d\n",ans[i]);
    }
    
    return 0;
}