#line 2 "/home/sakflat/CP/_library/cpp/template/template.cpp"

//yukicoder@cpp17

#include <iostream>
#include <iomanip>

#include <algorithm>

#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>

#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>

#include <bitset>

#include <complex>

#include <utility>

#include <numeric>

#include <functional>


using namespace std;
using ll = long long;
using P = pair<ll,ll>;

const ll MOD = 998244353;
const ll MODx = 1000000007;
const int INF = (1<<30)-1;
const ll LINF = (1LL<<62LL)-1;
const double EPS = (1e-10);

P ar4[4] = {{0,1},{0,-1},{1,0},{-1,0}};
P ar8[8] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};




template <typename T> vector<T> make_vector(size_t a, T b) { return vector<T>(a, b); }
template <typename... Ts> auto make_vector(size_t a, Ts... ts) { return vector<decltype(make_vector(ts...))>(a, make_vector(ts...)); }
 
 /*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる

計算量は変わらないが楽できるシリーズ
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる
*/

/*
function corner below
*/


/*
Function corner above
*/

/* comment outed because can cause bugs
__attribute__((constructor))
void initial() {
 cin.tie(0);
 ios::sync_with_stdio(false);
}
*/
long double f1(long double c, long double d){
  return min(c/3*4, d*4);
}

int main(){
  long double c,d;cin>>c>>d;
  long double mn = 0;
  long double mx = 1400;
  long long count = 0;
  while(count != 10000000){
    long double ce1 = (2*mn+mx)/3;
    long double ce2 = (mn+2*mx)/3;
    if(ce1*2/7 > c || ce1*5/7 > d){
      mx = ce1;
      continue;
    }
    if(ce2*2/7 > c || ce2*5/7 > d){
      mx = ce2;
      continue;
    }
    long double cost1 = ce1*2000 + f1(c-ce1*2/7, d-ce1*5/7)*1000; 
    long double cost2 = ce2*2000 + f1(c-ce2*2/7, d-ce2*5/7)*1000; 
    if(cost1 < cost2){
      mn = ce1;
    }else{
      mx = ce2;
    }
    count++;
  }
  cout << fixed << setprecision(10) << mx*2000 + f1(c-mx*2/7, d-mx*5/7)*1000 << endl;
  return 0;
}