Problem: problem

AOR Ika likes rooted trees with a fractal (self-similar) structure. Consider using the weighted rooted tree $ T $ consisting of $ N $ vertices to represent the rooted tree $ T'$ with the following fractal structure.

* $ T'$ is the addition of a tree rooted at $ x $ and having a tree structure similar to $ T $ (same cost) for each vertex $ x $ of $ T $.
* The root of $ T'$ is the same as that of $ T $.



The tree represented in this way is, for example, as shown in the figure below.

<image>

AOR Ika is trying to do a depth-first search for $ T'$, but finds that it takes a lot of time to trace all the vertices. Therefore, we decided to skip some node visits by performing a depth-first search with a policy of transitioning with a probability of $ p $ and not transitioning with a probability of $ 1-p $ at the time of transition during a depth-first search. Given $ T $ and the probability $ p $, find the expected value of the sum of the costs of all edges to follow when performing a depth-first search for $ T'$. The information of $ T $ is given by the number of vertices $ N $ and the information of the edges of $ N-1 $, and the vertex $ 1 $ is the root. Each vertex is labeled $ 1,2, \ dots, N $, and the $ i \ (1 \ le i \ le N-1) $ th edge costs the vertices $ x_i $ and $ y_i $ $ c_i $ It is tied with. The non-deterministic algorithm of the depth-first search that transitions to a child with a probability of $ p $, which is dealt with in this problem, is expressed as follows. The sum of the costs of the edges that the output $ \ mathrm {answer} $ follows.

1. Prepare an empty stack $ S $.
2. Set $ ​​\ mathrm {answer} = 0 $
3. Push the root vertex of $ T'$ to $ S $.
4. Extract the first element of $ S $ and make it $ x $.
5. For each child $ c $ of $ x $, do the following with probability $ p $ and do nothing with probability $ 1-p $.
* Add vertex $ c $ to $ S $. Then add the weight of the edge connecting $ x $ to $ c $ to $ \ mathrm {answer} $.
6. If S is not empty, transition to 3.
7. Output $ \ mathrm {answer} $.



output

Print the answer in one line. If the relative or absolute error is less than $ 10 ^ {-6} $, then AC. Also, output a line break at the end.

Example

Input

0.75
4
1 2 1
2 3 3
3 4 10


Output

24.8569335938

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<cstdio>
 
#define DEBUG
#define NMAX 100000
 
using namespace std;
 
typedef struct s{
    int y,c;
} S;
 
double p;
double AddC;
double sum2=0;
int N;

vector<S> v[NMAX+1];
bool f[NMAX+1];

double dfs(int n,double P,int C){
    //cout<<n<<","<<P<<","<<C<<endl;
    double sum=0;
    for(int i=0;i<v[n].size();i++){
        if(f[v[n][i].y]==false){
            f[v[n][i].y]=true;
            sum+=dfs(v[n][i].y,P*p,v[n][i].c);
        }
    }
    //cout<<"ret:"<<sum<<endl;
    return (double)C+sum*p;
}
 
double dfs2(int n,double P,int C){
    double sum=0;
    for(int i=0;i<v[n].size();i++){
        if(f[v[n][i].y]==false){
            f[v[n][i].y]=true;
            sum+=dfs2(v[n][i].y,P*p,v[n][i].c);
        }
    }
    return (double)C+sum*p+AddC;
}
 
int main(){
    int x,y,c;
 
    cin>>p;
    cin>>N;
     
    for(int i=0;i<N-1;i++){
        cin>>x>>y>>c;
        S SS;
        SS.y=y;
        SS.c=c;
        v[x].push_back(SS);
        SS.y=x;
        v[y].push_back(SS);
    }

    for(int i=0;i<=N;i++) f[i]=false;
    f[1]=true;
    AddC=dfs(1,1,0);
    //cout<<AddC<<endl;
    for(int i=0;i<=N;i++) f[i]=false;
    f[1]=true;
    printf("%.8f\n",dfs2(1,1,0));
}