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
c = map(int, raw_input().split())
if abs(c[2]-c[3]) > 1:
    print '-1' 
else:
    if c[2] == c[3]:
        if min(c[:2]) < c[2] or max(c[:2]) <= c[2]:
            print '-1'
        elif c[0] != c[2]:
            print '4'*(c[0]-c[2]-1)+'47'*c[2]+'7'*(c[1]-c[2])+'4'
        else:
            print '7'+'4'*(c[0]-c[2]+1)+'74'*(c[2]-1)+'7'*(c[1]-c[2])
    elif c[2] == c[3] + 1:
        if min(c[:2]) < c[2]:
            print '-1'
        else:
            print '4'*(c[0]-c[2])+'47'*c[2]+'7'*(c[1]-c[2])
    else:
        if min(c[:2]) < c[3]:
            print '-1'
        else:
            print '7'+'4'*(c[0]-c[2]-1)+'47'*c[2]+'7'*(c[1]-c[2]-1)+'4'
