#include <bits/stdc++.h>
using namespace std;

// #define int long long
#define rep(i, n) for (long long i = (long long)(0); i < (long long)(n); ++i)
#define reps(i, n) for (long long i = (long long)(1); i <= (long long)(n); ++i)
#define rrep(i, n) for (long long i = ((long long)(n)-1); i >= 0; i--)
#define rreps(i, n) for (long long i = ((long long)(n)); i > 0; i--)
#define irep(i, m, n) for (long long i = (long long)(m); i < (long long)(n); ++i)
#define ireps(i, m, n) for (long long i = (long long)(m); i <= (long long)(n); ++i)
#define irreps(i, m, n) for (long long i = ((long long)(n)-1); i > (long long)(m); ++i)
#define SORT(v, n) sort(v, v + n);
#define REVERSE(v, n) reverse(v, v+n);
#define vsort(v) sort(v.begin(), v.end());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) (long long)(x.size())

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vs = vector<string>;
using vpll = vector<pair<ll, ll>>;
using vtp = vector<tuple<ll,ll,ll>>;
using vb = vector<bool>;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

const ll INF = 1e9+10;
const ll MOD = 1e9+7;
const ll LINF = 1e18;

const ll maxn_eratosthenes = 1000005;
bool _is_prime[maxn_eratosthenes];
vector<ll> P;
vector<ll> min_prime_can_divide_i;

void eratosthenes(const ll N)
{
  P.clear();
  min_prime_can_divide_i = vll(maxn_eratosthenes+1, 1);
  for (ll i = 0; i <= N; i++) {
    _is_prime[i] = true;
  }
  _is_prime[0]=_is_prime[1]=false;
  for (ll i = 2; i <= N; i++) {
    if (_is_prime[i]) {
      for (ll j = 2 * i; j <= N; j += i) {
        _is_prime[j] = false;
        if(min_prime_can_divide_i[j]==1) min_prime_can_divide_i[j] = i;
      }
      P.emplace_back(i);
      min_prime_can_divide_i[i] = i;
    }
  }
}

signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  
  ll n; double p; cin>>n>>p;
  
  eratosthenes(maxn_eratosthenes-1);
  
  ll two[21];
  two[0]=1;
  rep(i,20) two[i+1]=two[i]*2;
  
  // min_prime_can_divide_i:=その数xを割ることができる最小の素数
  // min_prime_can_divide_iの素数の個数をmapで記録しつつ、now/min_prime_can_divide_i[now]へ遷移
  // 約数の数は素因数から求めるなら、例えば20=2^2*5^1 ... (2+1)*(1+1)=6 .. +1はその数を使わない通り数
  // この通り数から自身の数と1を除外するために-2する
  // 各i1個ずつをそれぞれ独立に見れば、sum(そのiが残る確率*1) = 全iを通して残る個数の期待値
  double ans=0;
  irep(i,2,n+1){
    ll now=i;
    map<ll,ll> m;
    while(min_prime_can_divide_i[now]!=1){
      m[min_prime_can_divide_i[now]]++;
      now/=min_prime_can_divide_i[now];
    }
    ll cnt=1;
    for(auto e: m){
      cnt*=e.second+1;
    }
    cnt-=2;
    ans+=powl((1-p), cnt);
    // cout<<cnt<<' ';
  }
  // cout<<endl;
  
  // irep(i,2,n+1){
  //   cout<<min_prime_can_divide_i[i]<<' ';
  // }
  // cout<<endl;
  
  printf("%.15lf\n", ans);
}