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>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define o(a) cout << a << endl
#define int long long
using namespace std;
typedef pair<int, int> P;

struct edge{
    int to, cost;
};

double p;
int n;
vector<edge> G[100010];
double x[100010];

void dfs(int pos, int prev){
    x[pos] = 0;
    for(int i = 0; i < G[pos].size(); i++){
        edge e = G[pos][i];
        if(e.to == prev) continue;
        dfs(e.to, pos);
        x[pos] += (double)(x[e.to] + e.cost) * p;
        cout << x[pos] << endl;
    }
}

double solve(int pos, int prev){
    double res = x[0];
    for(int i = 0; i < G[pos].size(); i++){
        edge e = G[pos][i];
        if(e.to == prev) continue;
        res += (solve(e.to, pos) + (double)e.cost) * p;
    }
    return res;
}

signed main(){
    cin >> p >> n;
    rep(i, 0, n - 1){
        int a, b, c;
        cin >> a >> b >> c;
        a--; b--;
        G[a]. push_back((edge){b, c});
        G[b]. push_back((edge){a, c});
    }
    dfs(0, -1);
    // rep(i, 0, n){
    //     printf("%.10lf\n", x[i]);
    // }
    printf("%.10lf\n", solve(0, -1));
}