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:
//???1:WA
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
#include <set>
#include <map>
#include <time.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


struct Info{
	Info(int arg_to,double arg_cost){
		to = arg_to;
		cost = arg_cost;
	}
	int to;
	double cost;
};

vector<Info> G[100001];
double table[100001],P,E = 0.0;

void recursive(int node_id,int parent){

	for(int i = 0; i < G[node_id].size(); i++){
		if(G[node_id][i].to == parent)continue;

		table[G[node_id][i].to] = table[node_id]*P;
		E += table[G[node_id][i].to]*G[node_id][i].cost;
		recursive(G[node_id][i].to,node_id);
	}
}

int main(){

	scanf("%lf",&P);

	int N;
	int from,to;

	scanf("%d",&N);

	double cost;

	for(int i = 0; i < N-1; i++){
		scanf("%d %d %lf",&from,&to,&cost);
		from--;
		to--;
		G[from].push_back(Info(to,cost));
		G[to].push_back(Info(from,cost));
	}

	table[0] = 1.0;
	recursive(0,-1);

	double ANS = E;
	for(int i = 0; i < N; i++)ANS += table[i]*E;

	printf("%.10lf\n",ANS);

	return 0;
}