Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya loves long lucky numbers very much. He is interested in the minimum lucky number d that meets some condition. Let cnt(x) be the number of occurrences of number x in number d as a substring. For example, if d = 747747, then cnt(4) = 2, cnt(7) = 4, cnt(47) = 2, cnt(74) = 2. Petya wants the following condition to fulfil simultaneously: cnt(4) = a1, cnt(7) = a2, cnt(47) = a3, cnt(74) = a4. Petya is not interested in the occurrences of other numbers. Help him cope with this task.

Input

The single line contains four integers a1, a2, a3 and a4 (1 ≤ a1, a2, a3, a4 ≤ 106).

Output

On the single line print without leading zeroes the answer to the problem — the minimum lucky number d such, that cnt(4) = a1, cnt(7) = a2, cnt(47) = a3, cnt(74) = a4. If such number does not exist, print the single number "-1" (without the quotes).

Examples

Input

2 2 1 1


Output

4774


Input

4 7 3 1


Output

-1
#include <bits/stdc++.h>
using namespace std;
int A1, A2, A3, A4;
int main() {
  int i;
  scanf("%d%d%d%d", &A1, &A2, &A3, &A4);
  if (abs(A3 - A4) > 1)
    printf("-1");
  else {
    if (A3 == A4) {
      if ((((A1) > (A2) ? (A1) : (A2)) - A3 - 1 < 0) ||
          (((A1) < (A2) ? (A1) : (A2)) < A3))
        printf("-1");
      else {
        if (A1 - A3 - 1 < 0) {
          for (i = 1; i <= A3; ++i) printf("74");
          for (i = A2 - A3; i >= 1; --i) printf("7");
        } else {
          for (i = A1 - A3 - 1; i >= 1; --i) printf("4");
          for (i = 1; i <= A3; ++i) printf("47");
          for (i = A2 - A3; i >= 1; --i) printf("7");
          printf("4");
        }
      }
    } else {
      if (A3 > A4) {
        if ((A1 - A3 < 0) || (A2 - A3 < 0))
          printf("-1");
        else {
          for (i = A1 - A3; i >= 1; --i) printf("4");
          for (i = 1; i <= A3; ++i) printf("47");
          for (i = A2 - A3; i >= 1; --i) printf("7");
        }
      } else {
        if ((A1 - A3 - 1 < 0) || (A4 - 2 < 0) || (A2 - A3 < 0))
          printf("-1");
        else {
          printf("74");
          for (i = A1 - A3 - 1; i >= 1; --i) printf("4");
          for (i = A4 - 2; i >= 1; --i) printf("74");
          for (i = A2 - A3; i >= 1; --i) printf("7");
          printf("4");
        }
      }
    }
  }
  printf("\n");
  return 0;
}
