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:
#define REP(i,n) for(int i=0; i<(int)(n); i++)

#include <cstdio>
inline int getInt(){ int s; scanf("%d", &s); return s; }
inline double getDouble(){ double s; scanf("%lf", &s); return s; }

#include <queue>
#include <set>

using namespace std;

vector<pair<int, int> > g[100000];
double pp[100000];
double p;

double dfs(int pos, double prob = 1, int prev = -1){
  ::pp[pos] = prob;

  double ans = 0;
  REP(i,g[pos].size()){
    const int next = g[pos][i].first;
    const int c = g[pos][i].second;

    if(next != prev){
      ans += p * (c + dfs(next , prob * p, pos));
    }
  }

  return ans;
}

int main(){
  ::p = getDouble();
  const int n = getInt();

  REP(i,n-1){
    const int a = getInt() - 1;
    const int b = getInt() - 1;
    const int c = getInt();

    g[a].push_back(make_pair(b, c));
    g[b].push_back(make_pair(a, c));
  }

  const double t = dfs(0);

  double ans = t;
  REP(i,n){
    ans += pp[i] * t;
  }

  printf("%.10f\n", ans);
}