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:
//T?????????????????????????±?????????????e??????????¢???????p[e]??¨????????¨, ?£cost[e] * p[e]????????????????????????
//T'?????????????????????????±?????????????????????¨????°??????§, T???????????????????????? + ?£(cost[e] + T????????????????????????) * p[e]????±????????????????
//??????, p[e] = p^(e??????????????????????????±??????????????§?????????)?????????????????£???, ??????????????±???(?????????????????°)???dfs??§?±?????????°?????????

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <tuple>
#include <cmath>
using namespace std;
typedef tuple<int, int, int> T;

double P;
int n;
vector<int> et[100000];
vector<T> edges;
int depth[100000];

void dfs(int p, int v, int dep = 0) {
	depth[v] = dep;
	for (int i = 0; i < et[v].size(); i++) {
		if (et[v][i] == p) continue;
		dfs(v, et[v][i], dep + 1);
	}
}

int main() {
	cin >> P >> n;
	for (int i = 0; i < n - 1; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		a--;
		b--;
		et[a].push_back(b);
		et[b].push_back(a);
		edges.push_back(T(a, b, c));
	}
	dfs(-1, 0);
	
	double expect = 0;
	static double p[100000];
	
	for (int i = 0; i < edges.size(); i++) {
		p[i] = pow(P, max(depth[get<0>(edges[i])], depth[get<1>(edges[i])]));
		expect += get<2>(edges[i]) * p[i];
	}
	
	double ans = expect;
	for (int i = 0; i < edges.size(); i++) {
		ans += (get<2>(edges[i]) + expect) * p[i];
	}
	
	printf("%.14f\n", ans);
	return 0;
}