#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxx=1e5+2;
struct  S
{
    int to,clo,val;
};
vector<S> a[maxx];
int deep[maxx],f[maxx][30],lg[maxx];
int len[maxx];
void dfs(int x,int fa){
    deep[x]=deep[fa]+1;
    f[x][0]=fa;
    for(int i=1;(1<<i)<=deep[x];i++){
        f[x][i]=f[f[x][i-1]][i-1];
    }
    for(auto t : a[x]){
        if(t.to==fa) continue;
        len[t.to]=len[x]+t.val;
        dfs(t.to,x);
    }
}
int lca(int x,int y){
    if(deep[x]<deep[y]) swap(x,y);
    while(deep[x]>deep[y]){
        x=f[x][lg[deep[x]-deep[y]]-1];
    }
    if(x==y) return x;
    for(int i=lg[deep[x]]-1;i>=0;i--){
        if(f[x][i]!=f[y][i])
            x=f[x][i],y=f[y][i];
    }
    return f[x][0];
}
struct node
{
    int clo,val,u,v;
};
node b[maxx];
struct D
{
    int id,clo,val;
};
vector<D> c[maxx];
int cnt[maxx],Len[maxx];
int tot[maxx],ans[maxx];
void DFS(int x,int fa){
    for(auto it : c[x]){
        tot[it.id]+=it.val*cnt[it.clo];
        ans[it.id]-=it.val*Len[it.clo];
    }
    for(auto t : a[x]){
        if(t.to==fa) continue;
        cnt[t.clo]++;
        Len[t.clo]+=t.val;
        DFS(t.to,x);
        cnt[t.clo]--;
        Len[t.clo]-=t.val;
    }
}
int main(){
    int n,m; cin>>n>>m;
    for(int i=1;i<n;i++){
        int x,y,c,w;
        scanf("%d%d%d%d",&x,&y,&c,&w);
        a[x].push_back(S{y,c,w});
        a[y].push_back(S{x,c,w});
    }
    dfs(1,0);
    for(int i=1;i<=n;i++)
      lg[i]=lg[i-1]+(1<<lg[i-1]==i);
    for(int i=1;i<=m;i++){
        scanf("%d%d%d%d",&b[i].clo,&b[i].val,&b[i].u,&b[i].v);
        int lc=lca(b[i].u,b[i].v);
        ans[i]=len[b[i].u]+len[b[i].v]-2*len[lc];
        c[b[i].u].push_back(D{i,b[i].clo,1});
        c[b[i].v].push_back(D{i,b[i].clo,1});
        c[lc].push_back(D{i,b[i].clo,-2});
    }
    DFS(1,0);
    for(int i=1;i<=m;i++){
        ans[i]+=tot[i]*b[i].val;
        printf("%d\n",ans[i]);
    }
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
typedef long long ll;
const int maxx=1e5+2;
struct  S
{
    int to,clo,val;
};
vector<S> a[maxx];
int deep[maxx],f[maxx][30],lg[maxx];
int len[maxx];
void dfs(int x,int fa){
    deep[x]=deep[fa]+1;
    f[x][0]=fa;
    for(int i=1;(1<<i)<=deep[x];i++){
        f[x][i]=f[f[x][i-1]][i-1];
    }
    for(auto t : a[x]){
        if(t.to==fa) continue;
        len[t.to]=len[x]+t.val;
        dfs(t.to,x);
    }
}
int lca(int x,int y){
    if(deep[x]<deep[y]) swap(x,y);
    while(deep[x]>deep[y]){
        x=f[x][lg[deep[x]-deep[y]]-1];
    }
    if(x==y) return x;
    for(int i=lg[deep[x]]-1;i>=0;i--){
        if(f[x][i]!=f[y][i])
            x=f[x][i],y=f[y][i];
    }
    return f[x][0];
}
struct node
{
    int clo,val,u,v;
};
node b[maxx];
struct D
{
    int id,clo,val;
};
vector<D> c[maxx];
int cnt[maxx],Len[maxx];
int tot[maxx],ans[maxx];
void DFS(int x,int fa){
    for(auto it : c[x]){
        tot[it.id]+=it.val*cnt[it.clo];
        ans[it.id]-=it.val*Len[it.clo];
    }
    for(auto t : a[x]){
        if(t.to==fa) continue;
        cnt[t.clo]++;
        Len[t.clo]+=t.val;
        DFS(t.to,x);
        cnt[t.clo]--;
        Len[t.clo]-=t.val;
    }
}
int main(){
    int n,m; cin>>n>>m;
    for(int i=1;i<n;i++){
        int x,y,c,w;
        scanf("%d%d%d%d",&x,&y,&c,&w);
        a[x].push_back(S{y,c,w});
        a[y].push_back(S{x,c,w});
    }
    dfs(1,0);
    for(int i=1;i<=n;i++)
      lg[i]=lg[i-1]+(1<<lg[i-1]==i);
    for(int i=1;i<=m;i++){
        scanf("%d%d%d%d",&b[i].clo,&b[i].val,&b[i].u,&b[i].v);
        int lc=lca(b[i].u,b[i].v);
        ans[i]=len[b[i].u]+len[b[i].v]-2*len[lc];
        c[b[i].u].push_back(D{i,b[i].clo,1});
        c[b[i].v].push_back(D{i,b[i].clo,1});
        c[lc].push_back(D{i,b[i].clo,-2});
    }
    DFS(1,0);
    for(int i=1;i<=m;i++){
        ans[i]+=tot[i]*b[i].val;
        printf("%d\n",ans[i]);
    }
}
```