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_N = 2*(int)1e5 + 5;

struct Graph{
public:
	int to;
	int cost;
	int depth;
};

struct XYC{
public:
	int x, y, c;

	XYC(){}

	void setX(int _x){
		x = _x;
	}

	void setY(int _y){
		y = _y;
	}

	void setC(int _c){
		c = _c;
	}

	int getX(void){
		return x;
	}

	int getY(void){
		return y;
	}

	int getC(void){
		return c;
	}
		
	bool operator<(const XYC &xyc) const{
		return (x != xyc.x ? x < xyc.x : y < xyc.y);
	}
};

vector<Graph> G[MAX_N];

int sumCost[MAX_N]; // ?????±???????????????????????????????¨?
int numNode[MAX_N]; // ?????±?????????????????????????????°
int maxDepth; 	// ?????§??±???

bool used[MAX_N];

void dfs(int curr, int depth, int prev){
	if (used[curr]) return;
	used[curr] |= true;

	for (int i = 0; i < G[curr].size(); ++i){
		int next = G[curr][i].to;
		if (!used[next] && next != prev){
			G[curr][i].depth = depth;
			maxDepth = max(maxDepth, depth);
			dfs(next, depth + 1, curr);
		} // end if
	} // end for
}

void dfs2(int curr, int depth, int prev){
	if (used[curr]) return;
	used[curr] |= true;	
	
	for (int i = 0;i < G[curr].size(); ++i){
		int next = G[curr][i].to;
		if (!used[next] && next != prev){
			sumCost[depth] += G[curr][i].cost;
			++numNode[depth]; 
			dfs2(next, depth + 1, curr);
		} // edn if
	} // end for
}

int main()
{
	memset(sumCost, 0., sizeof(sumCost));
	memset(numNode, 0, sizeof(numNode));
	maxDepth = 0;

	rep (i, MAX_N) G[i].clear();
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	double p; cin >> p;
	int N; cin >> N;
	vector<XYC> xyc(N-1);

	rep (i, N-1){
		int x, y, c;
		cin >> x >> y >> c;
		--x, --y;
		xyc[i].setX(x);
		xyc[i].setY(y);
		xyc[i].setC(c);
	} // end rep

	sort(ALL(xyc));

	rep (i, N-1){
		G[xyc[i].getX()].push_back((Graph){xyc[i].getY(), xyc[i].getC(), 0});
		G[xyc[i].getY()].push_back((Graph){xyc[i].getX(), xyc[i].getC(), 0});
	} // end rep

	memset(used, false, sizeof(used));	
	dfs(0, 1, -1);
	memset(used, false, sizeof(used));	
	dfs2(0, 1, -1);

	double orig = 0.;
	double currp = p;
	for (int i = 1; i <= maxDepth; ++i){
		orig += (double)sumCost[i] * currp;
		currp *= p;
	} // end for

	double res = orig;
//	cerr << "orig: " << orig << endl;
	currp = p;
	for (int k = 1; numNode[k] != 0; ++k){
		res += sumCost[k] * currp;
		res += numNode[k] * currp * orig;
		currp *= p;
//		cerr << "numNode: " << numNode[k] << " sumCost: " << sumCost[k] << endl;
	} // end for

	printf("%.8lf\n", res);

	return 0;
}