def decentNumber(n):
    # Write your code here
    
    t = n - n % 3
    found = False
    
    while t >= 0 and not found:
        if (n - t) % 5 == 0:
            found = True
        else:
            t = t - 3

    print("5" * t + "3" * (n-t)) if found else print(-1)

public static void decentNumber(int n) {
  // Start by finding the largest multiple of 3 less than or equal to n
  for (int i = n - (n % 3); i >= 0; i -= 3) {
    int remaining = n - i;

    // Check if the remaining is a multiple of 5
    if (remaining % 5 == 0) {
      // Append '5' i times
      StringBuilder result = new StringBuilder();
      result.append("5".repeat(i));

      // Append '3' (n - i) times
      result.append("3".repeat(remaining));

      System.out.println(result);
      return; // Exit as we've found the solution
    }
  }

  // If no valid number is found, print -1
  System.out.println(-1);
}

void decentNumber(int n) {
    int three = ((3 - (n % 3)) % 3 ) * 5;
    if(three > n) cout << -1;
    else cout << string(n - three, '5') << string(three, '3');
    cout << endl;
}

def decentNumber(n):
    for i in range(n + 1):
        if (n - i) % 3 == 0 and i % 5 == 0:
            print((n - i)*'5' + (i)*'3')
            return
    print(-1)

def decomposeNumberBy3and5(n):
    """If possible, get {3: a, 5: b}, such that n = a*3 + b*5.
    3 has the priority."""
    if n in (1, 2, 4, 7):
        return None

    match n % 3:
        case 0:
            return {3: n // 3, 5: 0}
        case 1:
            return {3: n // 3 - 3, 5: 2}  # 1 = 5 * 2 - 3 * 3
        case 2:
            return {3: n // 3 - 1, 5: 1}  # 2 = 5 * 1 - 3 * 1


def decentNumber(n):
    if (decomp := decomposeNumberBy3and5(n)) is None:
        print(-1)
    else:
        print("5" * decomp[3]*3 + "3" * decomp[5]*5)