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;
/*{{{*/  //template
#define REP(i,n) for(int i=0;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<29
#define LINF LLONG_MAX/3
#define MP make_pair
#define PB push_back
#define pb push_back
#define EB emplace_back
#define ALL(v) (v).begin(),(v).end()
#define all(v) ALL(v)
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
constexpr double eps = 1e-14; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/

double p;
constexpr int MN = 100001;
int N;
vector<pii> g[MN];

double poi;

double dfs1(int cur,int pre){
    double ret=0;
//    for(int i=0;i<g[cur].size();i++){ <- ok
    for(pii v : g[cur]){
        if(v.first==pre) continue;
        ret += p*(double)(v.second+dfs1(v.first,cur));
    }
    return ret;
}

double dfs2(int cur,int pre){
    double ret=poi;
    for(int i=0;i<g[cur].size();i++){
        if(g[cur][i].first==pre) continue;
        ret += p*(g[cur][i].second+dfs2(g[cur][i].first,cur));
    }
    return ret;
}

int main(){
    cin >> p;
    cin >> N;
    rep(i,N-1){
        int x,y,c;cin>>x>>y>>c;
        x--;y--;
        g[x].pb(MP(y,c));
        g[y].pb(MP(x,c));
    }
    poi = dfs1(0,-1);
    cout << Double(dfs2(0,-1)) << endl;
}