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<(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define each(a,x) for(auto a : (x))
#define all(a) (a).begin(),(a).end()
#define chmin(a,b) ((a) = min((a),(b)))
#define chmax(a,b) ((a) = max((a),(b)))
#define in_range(x,l,r) ((l)<=(x) && (x)<(r))
#define printvec(a) rep(i,a) cout << a[i] << " \n"[i+1==(a).size()];
#define fs first
#define sc second
#define em emplace
#define eb emplace_back
#define sz size()
#define MP make_pair
using namespace std;
typedef long long ll;
typedef double D;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vs;

const ll INF = 1e8;
const D EPS = 1e-8;
const ll MOD = 1e9+7;

long double dfs(int v, int par, int d, long double p,
                vector<int> &cnt, const vector< vector< pair<int,int> > > &t){
  cnt[d]++;
  long double res = 0;
  for(int i=0;i<t[v].size();i++){
    pair<int,int> e = t[v][i];
    int u = e.first, c = e.second;
    if(u==par) continue;
    res += (dfs(u,v,d+1,p,cnt,t) + c) * p;
  }
  return res;
}

int main(){
  long double p; cin >> p;
  int n; cin >> n;
  vector< vector< pair<int,int> > > t(n);
  rep(i,n-1){
    int a,b,c;
    cin >> a >> b >> c; a--; b--;
    t[a].push_back(MP(b,c));
    t[b].push_back(MP(a,c));
  }
  vector<int> cnt(n,0);
  cnt[0]++;
  long double exp = dfs(0,-1,0,p,cnt,t);

  long double ans = 0, k = 1;
  rep(i,n){
    ans += exp*cnt[i]*k;
    k *= p;
  }
  cout << fixed << setprecision(9) << ans << endl;
}