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 <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
using namespace std;

const int N = 123456;
typedef vector<int> VI;
typedef pair<int, int> PI;
vector<PI> edges[N];

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

double p;


double dfs(int v, int par) {
	double sum = 0;
	rep(i, edges[v].size()) {
		PI w = edges[v][i];
		if (par == w.first) continue;
		double res = dfs(w.first, v) + w.second;
		sum += res * p;
	}
	return sum;
}

double dfs2(int v, int par, double single) {
	double sum = single;
	rep(i, edges[v].size()) {
		PI w = edges[v][i];
		if (par == w.first) continue;
		double res = dfs2(w.first, v, single) + w.second;
		sum += res * p;
	}
	return sum;
}


// RUPC C
int main() {
	int n;
	cin >> p >> n;
	rep(i, n - 1) {
		int x, y, c;
		cin >> x >> y >> c;
		x--, y--;
		edges[x].push_back(PI(y, c));
		edges[y].push_back(PI(x, c));
	}
	double single = dfs(0, -1); // T
	double ret = dfs2(0, -1, single);
	printf("%.15f\n", ret);
	// your code goes here
	return 0;
}