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:
/*  -*- coding: utf-8 -*-
 *
 * 2807.cc: Fractal Tree
 */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

const int MAX_N = 100000;

/* typedef */

typedef pair<int,int> pii;
typedef vector<pii> vpii;

struct Stat {
  int i, prt;
  double p;
  Stat() {}
  Stat(int _i, int _prt, double _p): i(_i), prt(_prt), p(_p) {}
};

/* global variables */

vpii nbrs[MAX_N];

/* subroutines */

/* main */

int main() {
  double p;
  int n;
  scanf("%lf%d", &p, &n);

  for (int i = 1; i < n; i++) {
    int x, y, c;
    scanf("%d%d%d", &x, &y, &c);
    x--, y--;
    nbrs[x].push_back(pii(y, c));
    nbrs[y].push_back(pii(x, c));
  }

  queue<Stat> q;
  q.push(Stat(0, -1, 1.0));
  double t = 0.0;

  while (! q.empty()) {
    Stat u = q.front(); q.pop();
    vpii nbru = nbrs[u.i];
    double vp = u.p * p;

    for (vpii::iterator vit = nbru.begin(); vit != nbru.end(); vit++) {
      int &vi = vit->first, &vc = vit->second;
      if (vi != u.prt) {
	t += vc * vp;
	q.push(Stat(vi, u.i, vp));
      }
    }
  }
  //printf("t=%lf\n", t);

  q.push(Stat(0, -1, 1.0));
  double ans = t;
  
  while (! q.empty()) {
    Stat u = q.front(); q.pop();
    vpii nbru = nbrs[u.i];
    double vp = u.p * p;
    ans += t * u.p;

    for (vpii::iterator vit = nbru.begin(); vit != nbru.end(); vit++) {
      int &vi = vit->first;
      if (vi != u.prt)
	q.push(Stat(vi, u.i, vp));
    }
  }
  printf("%.10lf\n", ans);

  return 0;
}