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;

typedef long long ll;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define each(itr,c) for(__typeof(c.begin()) itr=c.begin(); itr!=c.end(); ++itr)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second

const int N=100000;

struct edge{int to,cost;};
vector<edge> G[N];

int c[N]={}, d[N]={};
double pw[N];

void dfs(int now, int par, int depth)
{
    d[now] = depth;
    for(edge &e:G[now])
    {
        if(e.to == par) continue;
        
        c[e.to] = e.cost;
        dfs(e.to,now,depth+1);
    }
}

int main()
{
    double p;
    int n;
    scanf(" %lf %d", &p, &n);
    rep(i,n-1)
    {
        int a,b,x;
        scanf(" %d %d %d", &a, &b ,&x);
        --a;
        --b;
        G[a].pb(edge{b,x});
        G[b].pb(edge{a,x});
    }

    dfs(0,-1,0);

    pw[0]=1;
    for(int i=1; i<N; ++i) pw[i]=pw[i-1]*p;

    double A=0,B=1;
    rep(i,n)
    {
        A += c[i]*pw[d[i]];
        B += pw[d[i]];
    }

    printf("%.10f\n", A*B);
    return 0;
}