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<ll, ll> P;

#define EACH(i,a) for (auto& i : a)
#define FOR(i,a,b) for (ll i = (a); i < (b); ++i)
#define RFOR(i,a,b) for (ll i=(b)-1; i>=a;i--)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define pb push_back
#define ALL(a) (a).begin(),(a).end()

const ll linf = 1e18;
const int inf = 1e9;
const double eps = 1e-12;
const double pi = acos(-1);

template<typename T>
istream& operator>>(istream& is, vector<T>& vec) {
    EACH(x,vec) is >> x;
    return is;
}
template<typename T>
ostream& operator<<(ostream& os, vector<T>& vec) {
    REP(i,vec.size()) {
        if (i) os << " ";
        os << vec[i];
    }
    return os;
}
template<typename T>
ostream& operator<<(ostream& os, vector<vector<T> >& vec) {
    REP(i,vec.size()) {
        if(i) os << endl;
        os << vec[i];
    }
    return os;
}

struct Edge {
    int to, cost;
};

double per;
double C;
vector< vector<Edge> > G;

double dp1(int v, int prev) {
    double res = 0;
    EACH(e, G[v]) {
        if (e.to == prev) continue;
        res += per * (e.cost + dp1(e.to, v));
    }
    return res;
}

double dp2(int v, int prev) {
    double res = C;
    EACH(e, G[v]) {
        if (e.to == prev) continue;
        res += per * (e.cost + dp2(e.to, v));
    }
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> per;
    int n; cin >> n;
    G.resize(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});
    }
    C = dp1(0, -1);
    cout << fixed << setprecision(10) << dp2(0, -1) << endl;
}