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>
using namespace std;
typedef long long ll;
typedef pair<int,double> PII;
const int inf = numeric_limits<int>::max();

double p;
int n;
vector< vector<PII> > G;
double dp[100100];
double dp2[100100];
int visited[100100];

double calc_dp(int node){
  visited[node] = 1;
  double ret = 0;
  for(int i = 0;i < (int)G[node].size();++i){
    if(!visited[G[node][i].first]){
      ret += p*(G[node][i].second
                + calc_dp(G[node][i].first));
    }
  }
  return dp[node] = ret;
}

double calc_dp2(int node){
  visited[node] = 1;
  double ret = dp[0];
  for(int i = 0;i < (int)G[node].size();++i){
    if(!visited[G[node][i].first]){
      ret += 
        p*(G[node][i].second + calc_dp2(G[node][i].first));
    }
  }
  return dp2[node] = ret;
}

int main(void){
  cin >> p >> n;
  G.resize(n);
  for(int i = 0;i < n-1;++i){
    int x,y;
    double c;
    cin >> x >> y >> c;
    --x;--y;
    G[x].push_back(PII(y,c));
    G[y].push_back(PII(x,c));
  }
  fill(dp,dp+100100,0);
  fill(dp2,dp2+100100,0);
  fill(visited,visited+100100,0);
  calc_dp(0);
  fill(visited,visited+100100,0);
  calc_dp2(0);
  cout << setprecision(15) << dp2[0] << endl;
  return 0;
}