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
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

#define CIN_ONLY if(1)
struct cww{cww(){
    CIN_ONLY{
        ios::sync_with_stdio(false);cin.tie(0);
        cout<<fixed;
    }
}}star;
#define fin "\n"
#define FOR(i,bg,ed) for(int i=(bg);i<(ed);i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second
#define pb push_back
#define DEBUG if(0)
#define REC(ret, ...) std::function<ret (__VA_ARGS__)>
template <typename T>inline void chmin(T &l,T r){l=min(l,r);}
template <typename T>inline void chmax(T &l,T r){l=max(l,r);}
template <typename T>
istream& operator>>(istream &is,vector<T> &v){
    for(auto &it:v)is>>it;
    return is;
}
typedef vector<int> V;
typedef vector<V> VV;
VV G;
int a[112345];
int b[112345];
double c[112345];
double f[112345];
double g[112345];
double P;
double dfs1(int v,int p){
    double& res=f[v];
    res=0;
    for(auto &e:G[v]){
            int u=a[e]^b[e]^v;
            if(u==p)continue;
            res+=dfs1(u,v)+c[e];
        }
    res*=P;
    return  res;
}
double dfs2(int v,int p){
    double &res=g[v];
    res=0;
    for(auto &e:G[v]){
            int u=a[e]^b[e]^v;
            if(u==p)continue;
            res+=dfs2(u,v)+c[e];
        }
    res*=P;
    res+=f[1];

    return  res;
}

int main(){
    cin>>P;
    int N;
    cin>>N;
    G.resize(N+1);
    REP(i,N-1){
        cin>>a[i]>>b[i]>>c[i];
        G[a[i]].pb(i);
        G[b[i]].pb(i);
    }
    dfs1(1,1);
    cout<<dfs2(1,1)<<endl;
    return 0;
}