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;

const double pi  = 2 * acos(0.0);
const double eps = 1e-8;

#define REP(i,a,b) for(int i=(a); i<(b);++i)
#define rep(i,n) REP(i,0,n)
#define INF (1<<29)
#define INFLL (1LL<<62)

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

int dx[8] = {0, 1, 0, -1, 1, -1, 1, -1};
int dy[8] = {1, 0, -1, 0, 1, -1, -1, 1};

struct Dst {
  int id;
  int cost;

  Dst(int id, int cost)
    : id(id), cost(cost) {}
};

struct Node {
  int id;
  vector<Dst> dsts;
};

double p;
int N;

double dfs(int now, int prev, double np, double Te, vector<Node> &G) {
  double e = 0.0;

  for (int i = 0; i < G[now].dsts.size(); i++) {
    if (G[now].dsts[i].id == prev)
      continue;
    e += (G[now].dsts[i].cost + Te) * np;
    e += dfs(G[now].dsts[i].id, now, np * p, Te, G);
  }

  return e;
}

int main(void) {
  ios_base::sync_with_stdio(false); cin.tie(0);
  cin >> p >> N;
  vector<Node> G(N);

  rep(i, N - 1) {
    int x, y, c;
    cin >> x >> y >> c;
    x--;
    y--;
    G[x].dsts.push_back(Dst(y, c));
    G[y].dsts.push_back(Dst(x, c));
  }

  double Te = dfs(0, -1, p, 0.0, G);
  double ans = dfs(0, -1, p, Te, G) + Te;

  cout << fixed << setprecision(10) << ans << endl;
  return 0; 
}