#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PI pair<int,int>
const int maxm=1e5+5;
struct EE{
    int v,c,d;
};
struct QQ{
    int x,y,op,id;
};
vector<EE>g[maxm];
vector<QQ>Q[maxm];
const int maxd=20;
int f[maxm][25];
int d[maxm];
int ans[maxm];
int n,q;
void dfs1(int x,int fa){
    for(int j=1;j<=maxd;j++){
        f[x][j]=f[f[x][j-1]][j-1];
    }
    for(auto i:g[x]){
        int v=i.v;
        if(v==fa)continue;
        d[v]=d[x]+1;
        f[v][0]=x;
        dfs1(v,x);
    }
}
int lca(int a,int b){
    if(d[a]<d[b])swap(a,b);
    for(int i=maxd;i>=0;i--){
        if(d[f[a][i]]>=d[b]){
            a=f[a][i];
        }
    }
    if(a==b)return a;
    for(int i=maxd;i>=0;i--){
        if(f[a][i]!=f[b][i]){
            a=f[a][i];
            b=f[b][i];
        }
    }
    return f[a][0];
}
//
int sum[maxm];
int cnt[maxm];
//
void dfs(int x,int fa,int dist){
    for(auto i:Q[x]){
        int x=i.x,y=i.y;
        ans[i.id]+=i.op*(dist-sum[x]+cnt[x]*y);
    }
    for(auto i:g[x]){
        int v=i.v;
        if(v==fa)continue;
        cnt[i.c]++;
        sum[i.c]+=i.d;
        dfs(v,x,dist+i.d);
        cnt[i.c]--;
        sum[i.c]-=i.d;
    }
}
signed main(){
    ios::sync_with_stdio(0);
    cin>>n>>q;
    for(int i=1;i<n;i++){
        int a,b,c,d;cin>>a>>b>>c>>d;
        g[a].push_back({b,c,d});
        g[b].push_back({a,c,d});
    }
    dfs1(1,1);
    for(int i=1;i<=q;i++){
        int x,y,u,v;cin>>x>>y>>u>>v;
        int lc=lca(u,v);
        Q[u].push_back({x,y,1,i});
        Q[v].push_back({x,y,1,i});
        Q[lc].push_back({x,y,-2,i});
    }
    dfs(1,1,0);
    for(int i=1;i<=q;i++){
        cout<<ans[i]<<endl;
    }
    return 0;
}
/*
在线做法:
树上距离可以用lca做,
边权替换操作只需要计算u和v之间有多少个颜色为x的边以及其边权和,
是在树上建可持久化线段树,直接维护就行了,
(或者树剖转序列再建树)

离线做法:
将询问拆分为u,v,lca三个点的询问,
每个询问查询当前点到根的距离-颜色为x的边权和+cnt[x]*y
颜色为x的边权和、cnt[x]直接用数组就可以维护了,
最后答案就是ans[u]+ans[v]-2*ans[lca].
*/

### Prompt

Please create a solution in CPP to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PI pair<int,int>
const int maxm=1e5+5;
struct EE{
    int v,c,d;
};
struct QQ{
    int x,y,op,id;
};
vector<EE>g[maxm];
vector<QQ>Q[maxm];
const int maxd=20;
int f[maxm][25];
int d[maxm];
int ans[maxm];
int n,q;
void dfs1(int x,int fa){
    for(int j=1;j<=maxd;j++){
        f[x][j]=f[f[x][j-1]][j-1];
    }
    for(auto i:g[x]){
        int v=i.v;
        if(v==fa)continue;
        d[v]=d[x]+1;
        f[v][0]=x;
        dfs1(v,x);
    }
}
int lca(int a,int b){
    if(d[a]<d[b])swap(a,b);
    for(int i=maxd;i>=0;i--){
        if(d[f[a][i]]>=d[b]){
            a=f[a][i];
        }
    }
    if(a==b)return a;
    for(int i=maxd;i>=0;i--){
        if(f[a][i]!=f[b][i]){
            a=f[a][i];
            b=f[b][i];
        }
    }
    return f[a][0];
}
//
int sum[maxm];
int cnt[maxm];
//
void dfs(int x,int fa,int dist){
    for(auto i:Q[x]){
        int x=i.x,y=i.y;
        ans[i.id]+=i.op*(dist-sum[x]+cnt[x]*y);
    }
    for(auto i:g[x]){
        int v=i.v;
        if(v==fa)continue;
        cnt[i.c]++;
        sum[i.c]+=i.d;
        dfs(v,x,dist+i.d);
        cnt[i.c]--;
        sum[i.c]-=i.d;
    }
}
signed main(){
    ios::sync_with_stdio(0);
    cin>>n>>q;
    for(int i=1;i<n;i++){
        int a,b,c,d;cin>>a>>b>>c>>d;
        g[a].push_back({b,c,d});
        g[b].push_back({a,c,d});
    }
    dfs1(1,1);
    for(int i=1;i<=q;i++){
        int x,y,u,v;cin>>x>>y>>u>>v;
        int lc=lca(u,v);
        Q[u].push_back({x,y,1,i});
        Q[v].push_back({x,y,1,i});
        Q[lc].push_back({x,y,-2,i});
    }
    dfs(1,1,0);
    for(int i=1;i<=q;i++){
        cout<<ans[i]<<endl;
    }
    return 0;
}
/*
在线做法:
树上距离可以用lca做,
边权替换操作只需要计算u和v之间有多少个颜色为x的边以及其边权和,
是在树上建可持久化线段树,直接维护就行了,
(或者树剖转序列再建树)

离线做法:
将询问拆分为u,v,lca三个点的询问,
每个询问查询当前点到根的距离-颜色为x的边权和+cnt[x]*y
颜色为x的边权和、cnt[x]直接用数组就可以维护了,
最后答案就是ans[u]+ans[v]-2*ans[lca].
*/

```