#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10,LOG=20;
int up[maxn][21],tin[maxn],tout[maxn],timer;
int n,T,dis[maxn];
struct edge{
    int u,col,w;
};
vector<edge> a[maxn];
void dfs(int nod,int pa){
    ++timer;
    tin[nod]=timer;
    up[nod][0]=pa;
    for(int i=1;i<=LOG;i++){
        up[nod][i]=up[up[nod][i-1]][i-1];
    }
    for(auto i:a[nod]){
        if(i.u!=pa){
            dis[i.u]=dis[nod]+i.w;
            dfs(i.u,nod);
        }
    }
    tout[nod]=++timer;
    return;
}
bool ancestor(int u,int v){
    return tin[u]<=tin[v]&&tout[u]>=tout[v];
}
int lca(int u,int v){
    if(ancestor(u,v))
        return u;
    if(ancestor(v,u))
        return v;
    for(int i=LOG;i>=0;i--){
        if(!ancestor(up[u][i],v))
            u=up[u][i];
    }
    return up[u][0];
}
int dp[maxn],ct[maxn],ans[maxn];
vector<int> q[maxn];
struct query{
    int col,w;
}b[maxn];
void dfs_2(int nod,int pa){
    for(int i:q[nod]){
        int f=(i<0?-2:1);
        i=abs(i);
        ans[i]+=(f*((ct[b[i].col]*b[i].w)-dp[b[i].col]+dis[nod]));
    }
    for(auto i:a[nod]){
        if(i.u!=pa){
            dp[i.col]+=i.w;
            ct[i.col]++;
            dfs_2(i.u,nod);
            dp[i.col]-=i.w;
            ct[i.col]--;
        }
    }
    return;
}

int main(){
    tout[0]=1e9;
    scanf("%d%d",&n,&T);
    int x,y,z,w;
    for(int i=1;i<n;i++){
        scanf("%d%d%d%d",&x,&y,&z,&w);
        a[x].push_back({y,z,w});
        a[y].push_back({x,z,w});
    }
    dfs(1,0);
    for(int i=1;i<=T;i++){
        scanf("%d%d%d%d",&b[i].col,&b[i].w,&x,&y);
        q[x].push_back(i);
        q[y].push_back(i);
        q[lca(x,y)].push_back(-i);
    }
    dfs_2(1,0);
    for(int i=1;i<=T;i++){
        printf("%d\n",ans[i]);
    }

}

### Prompt

Please formulate a cpp solution 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;
const int maxn=1e5+10,LOG=20;
int up[maxn][21],tin[maxn],tout[maxn],timer;
int n,T,dis[maxn];
struct edge{
    int u,col,w;
};
vector<edge> a[maxn];
void dfs(int nod,int pa){
    ++timer;
    tin[nod]=timer;
    up[nod][0]=pa;
    for(int i=1;i<=LOG;i++){
        up[nod][i]=up[up[nod][i-1]][i-1];
    }
    for(auto i:a[nod]){
        if(i.u!=pa){
            dis[i.u]=dis[nod]+i.w;
            dfs(i.u,nod);
        }
    }
    tout[nod]=++timer;
    return;
}
bool ancestor(int u,int v){
    return tin[u]<=tin[v]&&tout[u]>=tout[v];
}
int lca(int u,int v){
    if(ancestor(u,v))
        return u;
    if(ancestor(v,u))
        return v;
    for(int i=LOG;i>=0;i--){
        if(!ancestor(up[u][i],v))
            u=up[u][i];
    }
    return up[u][0];
}
int dp[maxn],ct[maxn],ans[maxn];
vector<int> q[maxn];
struct query{
    int col,w;
}b[maxn];
void dfs_2(int nod,int pa){
    for(int i:q[nod]){
        int f=(i<0?-2:1);
        i=abs(i);
        ans[i]+=(f*((ct[b[i].col]*b[i].w)-dp[b[i].col]+dis[nod]));
    }
    for(auto i:a[nod]){
        if(i.u!=pa){
            dp[i.col]+=i.w;
            ct[i.col]++;
            dfs_2(i.u,nod);
            dp[i.col]-=i.w;
            ct[i.col]--;
        }
    }
    return;
}

int main(){
    tout[0]=1e9;
    scanf("%d%d",&n,&T);
    int x,y,z,w;
    for(int i=1;i<n;i++){
        scanf("%d%d%d%d",&x,&y,&z,&w);
        a[x].push_back({y,z,w});
        a[y].push_back({x,z,w});
    }
    dfs(1,0);
    for(int i=1;i<=T;i++){
        scanf("%d%d%d%d",&b[i].col,&b[i].w,&x,&y);
        q[x].push_back(i);
        q[y].push_back(i);
        q[lca(x,y)].push_back(-i);
    }
    dfs_2(1,0);
    for(int i=1;i<=T;i++){
        printf("%d\n",ans[i]);
    }

}

```