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;
 
#define INF 1001000100010001000
#define MOD 1000000007
#define EPS 1e-10
#define int long long
#define rep(i, N) for (int i = 0; i < N; i++)
#define Rep(i, N) for (int i = 1; i < N; i++)
#define For(i, a, b) for (int i = (a); i < (b); i++)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define i_i pair<int, int>
#define vi vector<int>
#define vvi vector<vi >
#define vb vector<bool>
#define vvb vector<vb >
#define vp vector< i_i >
#define all(a) (a).begin(), (a).end()
#define Int(x) int x; scanf("%lld", &x);
#define int2(x, y) int x, y; scanf("%lld %lld", &x, &y);
 
//int dxy[5] = {0, 1, 0, -1, 0};
// assign

double p, ans, ret;
int n;
vector< vector< i_i > > g;

int cnt;

void solve(int pos, int bef, double per);
void dfs(int pos, int bef, double per);
double calc(int pos, int bef, double per);

signed main()
{
    cin >> p >> n;
    g.assign(n+1, vector< i_i >());
    rep(i, n-1) {
        int2(a, b);
        Int(c);
        g[a].pb(mp(b, c));
        g[b].pb(mp(a, c));
    }

    solve(1, 0, p);
    //printf("%.10lf\n", ans);
    dfs(1, 0, 1.0);
    printf("%.10lf\n", ans + ret);

    //printf("%.10lf\n", ret);

    return 0;
}


void solve(int pos, int bef, double per)
{
    //cout << pos << endl;

    if (per < EPS) return ;
    for (i_i i : g[pos]) {
        if (i.first == bef) continue;
        solve(i.first, pos, per * p);
        ans += per * (double)i.second;
    }
}
        
void dfs(int pos, int bef, double per)
{
    if (per < EPS) return;
    ret += per * ans;
    for (i_i i : g[pos]) {
        if (i.first == bef) continue;
        dfs(i.first, pos, per * p);
    }
}

double calc(int pos, int bef, double per)
{
   // cout << cnt++ << endl;
    if (per < EPS) return 0.0;
    for (i_i i : g[pos]) {
        if (i.first == bef) continue;
        ans += per * i.second + calc(i.first, pos, per * p) * per;
    }
    return ans;
}