from collections import deque
before = input()
N = int(input())
after = input()

if before.count('o') != after.count('o'):
    exit(print('SUCCESS'))
# これ以降 o の個数は同じ

if before == 'ooo' or before == 'xxx':
    exit(print('FAILURE'))
# o1つかo2つのパターンのみ

if (before == 'oxo' or before == 'xox') and (before == after) and N == 1:
    exit(print('SUCCESS'))
# 最初のマスで時間稼ぎ可能

def convert(s:str):
    res = 0
    for i in range(3):
        if s[i] == 'o':
            res += 1<<(2 - i)
    return res

def revert(n:int):
    memo = format(n, 'b').zfill(3)
    res = []
    for i in range(3):
        if memo[i] == '1':
            res.append('o')
        else:
            res.append('x')
    return res

dist = [10]*8
start = convert(before)
goal = convert(after)

dist[start] = 0
que = deque([start])
while que:
    pos = que.popleft()

    # 頂点番号から文字列
    memo = revert(pos)

    # 2方向への遷移を検討
    memo1 = [memo[1], memo[0], memo[2]]
    nex1 = convert(''.join(memo1))
    if dist[nex1] == 10:
        dist[nex1] = dist[pos] + 1
        que.append(nex1)

    memo2 = [memo[0], memo[2], memo[1]]
    nex2 = convert(''.join(memo2))
    if dist[nex2] == 10:
        dist[nex2] = dist[pos] + 1
        que.append(nex2)

if N < dist[goal]:
    print('SUCCESS')
else:
    print('FAILURE')