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<bits/stdc++.h>
using namespace std;
struct edge{ int to,cost; };

double p;
int N;
vector<edge> G[100005];

double x[100005];

void dfs(int pos,int prev){
  x[pos]=0;
  for(int i=0;i<(int)G[pos].size();i++){
    edge e=G[pos][i];
    if(e.to==prev)continue;
    dfs(e.to,pos);
    x[pos]+=(double)(x[e.to]+e.cost)*p;
  }
}

double solve(int pos,int prev){
  double res=x[0];
  for(int i=0;i<(int)G[pos].size();i++){
    edge e=G[pos][i];
    if(e.to==prev)continue;
    res+=( solve(e.to,pos) + e.cost )*p;
  }
  return res;
}

int main(){
  cin>>p;
  cin>>N;
  for(int i=0;i<N-1;i++){
    int a,b,c;
    cin>>a>>b>>c;
    a--,b--;
    G[a].push_back( (edge){b,c} );
    G[b].push_back( (edge){a,c} );
  }
  dfs(0,-1);
  /*
  for(int id=0;id<N;id++){
    printf("%d %.10f\n", id,x[id]);
  }
  */
  printf("%.10f\n", solve(0,-1));
  return 0;
}