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;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

double q;
int N;


double ans0;
vpint G[111111];

double dfs(int v,int p){
    double res=0;
    for(auto &e:G[v]){
        if(e.fi==p)continue;
        double tmp=dfs(e.fi,v);
        res+=q*(e.se+tmp);
    }
    return res;
}

double dfs2(int v,int p){
    double res=ans0;
    for(auto &e:G[v]){
        if(e.fi==p)continue;
        double tmp=dfs2(e.fi,v);
        res+=q*(e.se+tmp);
    }
    return res;
}

signed main(){
    cin>>q>>N;
    rep(i,N-1){
        int a,b,c;
        cin>>a>>b>>c;
        a--;b--;
        G[a].pb({b,c});
        G[b].pb({a,c});
    }
    ans0=dfs(0,-1);
    printf("%.20f\n",dfs2(0,-1));
    return 0;
}