s2p = {'oox': 2, 'oxo': 1, 'xoo': 0,
       'xxo': 2, 'xox': 1, 'oxx': 0}

def solve(s_before, N, s_after):
    if s_before.count('o') != s_after.count('o'):
        return 'SUCCESS'
    if s_before.count('o') == 0 or s_before.count('o') == 3:
        return 'FAILURE'
    p0 = s2p[s_before]
    p1 = s2p[s_after]
    if N == 0 and p0 != p1:
        return 'SUCCESS'
    if N == 1:
        if abs(p0 - p1) == 2 or (p0 == 1 and p1 == 1):
            return 'SUCCESS'
    return 'FAILURE'

s_before = input().rstrip()
N = int(input())
s_after = input().rstrip()

print(solve(s_before, N, s_after))