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 <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <string>
#include <map>
#include <set>
#include <functional>
#include <iostream>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
double p;
class edge{
public:
	int t;
	double c;
	edge(){}
	edge(int tt,double cc){
		t=tt;
		c=cc;
	}
};

vector<edge> G[100001];
int depth[100001];
int cnt[100001];
double sum=0.0;
bool flag[100001];

void dfs(int v,int d){
	flag[v]=true;
	depth[v]=d;
	for(int i=0;i<G[v].size();i++){
		if(!flag[G[v][i].t]){
			sum+=G[v][i].c*pow(p,(d+1.0));
			dfs(G[v][i].t,d+1);
		}
	}
}

int main(void){
	scanf("%lf",&p);
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int f,t,c;
		scanf("%d%d%d",&f,&t,&c);
		f--;
		t--;
		G[f].push_back(edge(t,(double)c));
		G[t].push_back(edge(f,(double)c));
	}
	memset(flag,false,sizeof(flag));
	dfs(0,0);
	for(int i=0;i<n;i++){
		cnt[depth[i]]+=1;
	}
	cnt[0]++;
	double res=0.0;
	for(int i=0;i<n;i++){
		res+=(double)cnt[i]*sum*pow(p,i);
	}
	printf("%.9f\n",res);
	return 0;
}