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"
#include<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
using ld = long double;
const ld eps = 1e-9;

struct edge {
	int from;
	int to;
	int cost;
	edge() {
		from = 0;
		to = 0;
		cost = 0;
	}
	edge(int from_, int to_, int cost_):from(from_), to(to_), cost(cost_) {

	}
};
ld p;
ld dfs1(const int now, const int from, const vector<vector<edge>>&edges,vector<ld>&come_pers,ld aper) {
	ld ans = 0;
	come_pers[now] = aper;
	for (auto e : edges[now]) {
		if (e.to == from)continue;
		else {
			ans += p*(e.cost + dfs1(e.to, now, edges,come_pers,aper*p));
		}
	}
	return ans;
}

int main() {cin >> p;
	int N; cin >> N;
	vector<vector<edge>>edges(N);
	for (int i = 0; i < N - 1; ++i) {
		int x, y, c; cin >> x >> y >> c;
		x--; y--;
		edges[x].push_back(edge(x, y, c));
		edges[y].push_back(edge(y, x, c));
	}
	vector<ld>come_pers(N);
	ld sum=dfs1(0, -1, edges, come_pers, 1);
	ld ans = accumulate(come_pers.begin(), come_pers.end(), 1.l)*sum;
	cout << setprecision(10) << fixed << ans << endl;
	return 0;
}