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;

double p;
vector<vector<pair<int, int>>> e;
bool done[100010];

double value_dfs(int pos, double curr_p) {
    done[pos] = true;

    curr_p *= p;
    double ret = 0.;
    for (auto x : e[pos]) {
        int to = x.first;
        int cost = x.second;
        if (!done[to]) {
            ret += cost * curr_p;
            ret += value_dfs(to, curr_p);
        }
    }
    return ret;
}

double dfs2(int pos, double curr_p) {
    if (done[pos]) return 0.;
    done[pos] = true;

    double ret = curr_p;

    curr_p *= p;
    for (auto x : e[pos]) {
        int to = x.first;
        if (done[to]) continue;
        ret += dfs2(to, curr_p);
    }
    return ret;
}

int main() {
    int N;
    cin >> p >> N;
    e = vector<vector<pair<int, int>>>(N);
    for (int i = 0; i < N - 1; i++) {
        int x, y, c;
        cin >> x >> y >> c;
        x--, y--;
        e[x].push_back({y, c});
        e[y].push_back({x, c});
    }

    memset(done, false, sizeof(done));
    double v = value_dfs(0, 1.0);
    memset(done, false, sizeof(done));
    printf("%.10lf\n", v + v * dfs2(0, 1.0));
}