44 58 90 53
 20 14 37 71
 4 2 4 2
 1 2 5 5
 1 2 8 9
 99 100 3 98
 92 69 83 97
 3 1 1 1
 18 88 18 18
 1 1 1 1
 69 7 44 30
 90 30 30 30
 1 2 3 5
 1 1 3 1
 6 3 6 6
 26 52 7 19
 6 4 8 6
 10 2 3 5
 34 11 84 39
 1 1 14 12
 3 2 1 1
 2 4 6 6
 1 1 2 1
 72 96 2 26
 97 95 91 27
 1 2 1 3
 1 1 2 3
 1 1 1 2
 1 1 10 20
 2 3 2 5
 1 2 10 7
 2 1 28 9
 4 4 4 8
 2 2 4 6
 72 52 62 62
 76 97 99 74
 100 100 99 100
 30 74 41 63
 5 10 1 6
 18 17 17 20
 2 4 1 1
 7 3 6 3
 1 1 4 5
 2 2 6 2
 1 2 6 3
 3 5 1 3
 70 100 10 86
 48 14 3 31
 2 6 3 2
 2 3 10 10
 4 4 12 4
 14 9 10 6
 100 98 99 97
 5 7 1 3
 1 1 4 2
 14 10 18 24
 1 100 100 1
 1 2 4 5
 1 2 3 4
 1 2 3 3
 4 4 4 5
 66 68 16 82
 1 2 3 2
 2 3 3 4
 100 100 100 100
 1 2 2 7
 3 14 36 53
 1 1 3 4

NO
 YES
 YES
 NO
 YES
 NO
 NO
 YES
 NO
 YES
 NO
 YES
 NO
 YES
 NO
 YES
 YES
 YES
 YES
 YES
 NO
 NO
 NO
 YES
 NO
 NO
 NO
 NO
 NO
 NO
 YES
 NO
 NO
 NO
 YES
 YES
 NO
 YES
 YES
 NO
 YES
 NO
 NO
 YES
 YES
 YES
 NO
 YES
 NO
 NO
 YES
 NO
 YES
 YES
 YES
 NO
 YES
 YES
 YES
 NO
 NO
 NO
 YES
 YES
 YES
 NO
 YES
 NO

__author__ = 'Devesh Bajpai'

'''
https://codeforces.com/problemset/problem/1230/A

Solution: The cheap solution is to try out all the combinations possible for the two friends when we have 4 candy bags.
So the grouping can either be 2-2 or 1-3. Hence we brute-force to try all combinations in those groupings. 

It would be interesting to extend this question beyond 4 candy bags. 
'''


def solve(a1,a2,a3,a4):

    # Try combinations of 2,2
    if (a1 + a2 == a3 + a4) or (a1 + a3 == a2 + a4) or (a1 + a4 == a2 + a3) or (a2 + a3 == a2 + a4):
        return "YES"

    # Try combinations of 1,3
    elif (a1 + a2 + a3 == a4) or (a2 + a3 + a4 == a1) or (a1 + a3 + a4 == a2) or (a1 + a2 + a4 == a3):
        return "YES"

    else:
        return "NO"


if __name__ == "__main__":
    a1,a2,a3,a4 = map(int,raw_input().split(" "))

    print solve(a1,a2,a3,a4)