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++)
using namespace std;
typedef pair<int, int> pii;
typedef vector< vector<pii> > Graph;

struct Edge {
    int from, to, cost;
    Edge(int a, int b, int c) : from(a), to(b), cost(c) {};
};

int depth[100010];
void getDepth(Graph &G, int point) {
    for(size_t i=0; i<G[point].size(); i++) {
        int to = G[point][i].first;
        if(depth[to] != -1) continue;
        depth[to] = depth[point] + 1;
        getDepth(G, to);
    }
}

int main() {
    double p; cin >> p;
    int n; cin >> n;
    Graph G(n);
    vector<Edge> ve;
    rep(i,0,n-1) {
        int x, y, c;
        cin >> x >> y >> c;
        x--; y--;
        G[x].push_back(pii(y, c));
        G[y].push_back(pii(x, c));
        ve.push_back(Edge(x, y, c));
    }

    memset(depth, -1, sizeof(depth));
    depth[0] = 0;
    getDepth(G, 0);

    double E = 0.0;
    rep(i,0,ve.size()) {
        int x = depth[ ve[i].from ];
        int y = depth[ ve[i].to ];
        int d = max(x, y);
        E += ve[i].cost * pow(p, d);
    }

    double ans = 2 * E;
    rep(i,0,ve.size()) {
        int x = depth[ ve[i].from ];
        int y = depth[ ve[i].to ];
        int d = max(x, y);
        ans += E * pow(p, d);
    }

    cout << fixed << setprecision(10) << ans << endl;
}