//http://www.prefield.com/algorithm/math/two_stage_simplex.html
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>

using namespace std;

typedef vector<double> Array;
typedef vector<Array> Matrix;

const double EPS = 1e-8;
enum { OPTIMAL, UNBOUNDED, NOSOLUTION, UNKNOWN };
struct two_stage_simplex {
  int N, M, st;
  Matrix a;
  vector<int> s;
  two_stage_simplex(const Matrix &A, const Array &b, const Array &c)
    : N(A.size()), M(A[0].size()), a(N+2, Array(M+N+1)), s(N+2), st(UNKNOWN) {
    for (int j = 0; j < M; ++j) a[N+1][j] = c[j]; // make simplex table
    for (int i = 0; i < N; ++i)
      for (int j = 0; j < M; ++j) a[i+1][j] = A[i][j];
    for (int i = 0; i < N; ++i) a[i+1][M+N] = b[i]; // add helper table
    for (int i = 0; i < N; ++i) a[ 0 ][i+M] = 1;
    for (int i = 0; i < N; ++i) a[i+1][i+M] = 1;
    for (int i = 0; i < N; ++i) s[i+1]      = i+M;
    for (int i = 1; i <= N; ++i)
      for (int j = 0; j <= N+M; ++j) a[0][j] += a[i][j];
    st = solve();
  }
  int status() const { return st; }
  double solution() const { return -a[0][M]; }
  double solution(Array &x) const {
    x.resize(M, 0);
    for (int i = 0; i < N; ++i)
      x[s[i+1]] = a[i+1].back();
    return -a[0][M];
  }
  int solve() {
    M += N; N += 1;
    solve_sub(); // solve stage one
    if (solution() > EPS) return NOSOLUTION;
    N -= 1; M -= N;
    swap(a[0], a.back()); a.pop_back(); // modify table
    for (int i = 0; i <= N; ++i) {
      swap(a[i][M], a[i].back());
      a[i].resize(M+1);
    }
    return solve_sub(); // solve stage two
  }
  int solve_sub() {
    int p, q;
    while (1) {
      //print();
      for (q = 0; q <= M && a[0][q] >= -EPS; ++q);
      for (p = 0; p <= N && a[p][q] <=  EPS; ++p);
      if (q >= M || p > N) break;
      for (int i = p+1; i <= N; ++i) // bland's care for cyclation
        if (a[i][q] > EPS)
          if (a[i][M]/a[i][q] < a[p][M]/a[p][q] ||
             (a[i][M]/a[i][q] == a[p][M]/a[p][q] && s[i] < s[q])) p = i;
      pivot(p, q);
    }
    if (q >= M) return OPTIMAL;
    else        return UNBOUNDED;
  }
  void pivot(int p, int q) {
    for (int j = 0; j <= N; ++j)
      for (int k = M; k >= 0; --k)
        if (j != p && k != q)
          a[j][k] -= a[p][k]*a[j][q]/a[p][q];
    for (int j = 0; j <= N; ++j)
      if (j != p) a[j][q] = 0;
    for (int k = 0; k <= M; ++k)
      if (k != q) a[p][k] = a[p][k]/a[p][q];
    a[p][q] = 1.0;
    s[p] = q;
  }
};

int main() {
  /*for (int n, m; cin >> n >> m; ) {
    Array c(n+m), b(m);
    for (int i = 0; i < n; ++i)
      cin >> c[i], c[i] *= -1;
    Matrix A(m, Array(n+m));
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j)
        cin >> A[i][j];
      A[i][n+i] = 1;
      cin >> b[i];
    }
    two_stage_simplex tss(A, b, c);
    double ans = -tss.solution();
    printf("Nasa can spend %.0f taka.\n", ans + 0.5 - EPS);
  }*/
  
  //整数計画ソルバーが欲しい（これは実数で計画しているからWAだと思う）
	Array c(5), b(5);
	for (int i = 0; i < 5; i++) { c[i] = 1; c[i] *= -1; }
	Matrix A(5, Array(10));
	for (int i = 0; i < 5; i++) {
		for (int j = 0; j < 3; j++) {
			A[i][(i + j) % 5] = 1;
		}
		A[i][5 + i] = 1;
	}
	cin >> b[2] >> b[3] >> b[4] >> b[0] >> b[1];
	two_stage_simplex tss(A, b, c);
	double ans = -tss.solution();
	printf("%lld\n", (long long)(ans + 1e-9));
	return 0;
}