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;
typedef long long ll;

struct edge {
  int to, cost;
  edge() {}
  edge(int t, int c) : to(t), cost(c) {}
};

int n;
double p;
vector<edge> G[100010];
double dp1[100010], dp2[100010];

void f(int v, int pv) {
  for (int i = 0; i < G[v].size(); i++) {
    int ch = G[v][i].to;
    if (ch == pv) continue;
    f(ch, v);
    dp1[v] += G[v][i].cost + dp1[ch];
  }
  dp1[v] *= p;
}

void g(int v, int pv) {
  for (int i = 0; i < G[v].size(); i++) {
    int ch = G[v][i].to;
    if (ch == pv) continue;
    g(ch, v);
    dp2[v] += G[v][i].cost + dp2[ch];
  }
  dp2[v] *= p;
  dp2[v] += dp1[0];
}

int main() {
  cin >> p >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    x--, y--;
    G[x].push_back(edge(y, c));
    G[y].push_back(edge(x, c));
  }
  f(0, -1);
  g(0, -1);
  printf("%.10f\n", dp2[0]);

  return 0;
}