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;

const bool debug = true;
#define dbg(...) if(debug) printf(__VA_ARGS__)
#define print(var) if (debug) cout << #var << " = " << var << endl

namespace {
    using real = long double;
    struct Edge {
        int from, to, cost;
        Edge() {}
        Edge(int from, int to, int cost) : from(from), to(to), cost(cost) {}
    };
    real p;
    int N;
    vector<vector<Edge>> G;
    void input() {
        cin >> p >> N;
        G.resize(N);
        for (int i = 0; i < N - 1; i++) {
            int x, y, c; cin >> x >> y >> c;
            x--; y--;
            G[x].emplace_back(x, y, c);
            G[y].emplace_back(y, x, c);
        }
    }

    void solve() {
        real e; {
            function<real(int, int)> dfs = [&](int v, int prev) {
                // returns expected value of cost.:
                real ret = 0;
                for (auto& next : G[v]) {
                    if (next.to == prev) continue;
                    ret += p * (next.cost + dfs(next.to, v));
                }
                return ret;
            };
            e = dfs(0, -1);
        }
        real ans = 0; {
            function<void(int, int, real)> dfs = [&](int v, int prev, real prob) {
                ans += prob * e;
                for (auto& next : G[v]) {
                    if (next.to == prev) continue;
                    ans += prob * p * next.cost;
                    dfs(next.to, v, prob * p);
                }
            };
            dfs(0, -1, 1.0);
        }
        cout << setprecision(12) << fixed << ans << endl;
    }
}

int main() {
    input(); solve();
    return 0;
}