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>
#define rep(i,n) for(int i=0;i<(n);++i)
#define ALL(A) A.begin(), A.end()

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX_V = (int)1e5 + 1;

int n;
double p;

class edge{
public:
	int to;
	double cost;
	edge(){}
	edge(int _to, double _cost):to(_to),cost(_cost){}
};

vector<edge> G[MAX_V];
int depth[MAX_V];
int cnt[MAX_V];
double sum = 0.0;
bool visit[MAX_V];

void dfs(int v, int d){
	visit[v] |= true;
	depth[v] = d;
	for (int i = 0; i < (int)G[v].size(); ++i){
		if (!visit[G[v][i].to]){
			sum += G[v][i].cost * pow(p,(d+1.0));
			dfs(G[v][i].to, d+1);
		} // end if
	} // end for
}

int main(){
	
	// init
	for (int i = 0; i < MAX_V; ++i) G[i].clear();
	memset(depth, 0, sizeof(depth));
	memset(cnt, 0, sizeof(cnt));
	memset(visit, false, sizeof(visit));

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin >> p >> n;
	for (int i = 0; i < n - 1; ++i){
		int from, to, cost;
		cin >> from >> to >> cost;
		--from;
		--to;
		G[from].push_back(edge(to,(double)cost));
		G[to].push_back(edge(from,(double)cost));
	} // end for

	dfs(0,0);
	for (int i = 0; i < n; ++i){
		++cnt[depth[i]];
	} // end for
/*
	// debug
	for (int i = 0; i < n; ++i){
		cerr << "v: " << i << " depth: " << depth[i] << endl;
	} // end for
*/
	++cnt[0]; // ????????±??? 0 ????????§ ???????¨?????????????0???????????????????????????????????????

	double res = 0.0;
	for (int i = 0; i < n; ++i){
		res += (double)cnt[i] * sum * pow(p,i);
	} // end for
	cout << setprecision(9) << res << endl;

	return 0;
}