#include <bits/stdc++.h>
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define FORR(i,a,b) for (int i=(a);i>=(b);i--)
#define pb push_back
#define pcnt __builtin_popcount
#define show(x) cout<<#x<<" = "<<x<<endl;
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define fi first
#define se second
#define rng(a) a.begin(),a.end()
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define sz(x) (int)(x).size()

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef set<int> si;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<pll> vpll;
typedef set<ll> sl;
template<typename T>string join(vector<T>&v)
{stringstream s;FOR(i,0,sz(v))s<<' '<<v[i];return s.str().substr(1);}
ll gcd(ll a,ll b){if(a>b)swap(a,b);for(;a>0;b%=a,swap(a,b));return b;}
int modpow(ll a,ll n,int m){if(a==0)return a;ll p=1;for(;n>0;n/=2,a=a*a%m)if(n&1)p=p*a%m;return(int)p;}
void dout(double d){printf("%.12f\n",d);}

const int iinf = 1e9;
const ll linf = 1e18;
const int mod = 1e9+7;
const double pi = acos(-1);
const double eps = 1e-10;

double r[2], m[2], e[2][2] = {
  {0.75, 2./7},
  {0.25, 5./7}
};

const int GN=2;
struct Gauss{
  int solve(int n, double o[GN][GN], double u[GN], double r[GN]) {
    int q;
    double m[GN][GN],v[GN], p;
    memmove(m,o,sizeof(m));
    memmove(v,u,sizeof(v));
    FOR(i, 0, n) {
      p = 0.;
      q = 0;
      FOR(j, i, n) if(p < abs(m[i][j])){
        p = abs(m[i][j]);
        q = j;
      }
      if(p < eps) return -1;
      swap(v[i],v[q]);
      FOR(k, i, n) swap(m[i][k], m[q][k]);
      v[i]/=m[i][i];
      FORR(k, n-1, i) m[i][k]/=m[i][i];
      FOR(j, i+1, n){
        v[j]-=v[i]*m[j][i];
        FORR(k, n-1, i) m[j][k]-=m[i][k]*m[j][i];
      }
    }
    FORR(i, n-1, 0) {
      FORR(j, n-1, i+1) v[i]-=m[i][j]*v[j],m[i][j]=0;
      r[i]=v[i];
    }
    return 0;
  }
};
Gauss g;

main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> m[0] >> m[1];
  g.solve(2, e, m, r);
  double ans;
  if(r[0] >= 0 && r[1] >= 0){
    ans = r[0] * 1000 + r[1] * 2000;
  }else if(r[1] < 0){
    ans = m[1] * 4 * 1000;
  }else{
    ans = m[0] * 7 / 2 * 2000;
  }
  dout(ans);
}