# o, xの数が異なれば常にsuccess
# 数が同じ時
# 数が同じでもooo or xxxは常にfailure
# 数が同じで1個2個パターンのときどうなるか oxx, oox 
# 2回以上ならどんなパターンにもできるので、常にfailure
# 数が同じで、1個2個で、1回だけのときは、できるパターンが限られる、できるならsuccess, or failure

S_before = input()
count_before = S_before.count('o')
N = int(input())
S_after = input()
count_after = S_after.count('o')

if count_before != count_after:
    ans = 'SUCCESS'
else:
    # つまり数は同じ
    if count_before == 0 or count_before == 3:
        ans = 'FAILURE'
    else:
        # つまり数は同じで1個2個に分かれている
        if N >= 2: 
            ans = 'FAILURE'
        elif N == 0:
            if S_before == S_after:
                ans = 'FAILURE'
            else:
                ans = 'SUCCESS'
        elif N == 1:
            if S_before == 'oxx':
                if S_after == 'oxx' or S_after == 'xox':
                    ans = 'FAILURE'
                else:
                    ans = 'SUCCESS'
            if S_before == 'xox':
                if S_after == 'oxx' or S_after == 'xxo':
                    ans = 'FAILURE'
                else:
                    ans = 'SUCCESS'            
            if S_before == 'xxo':
                if S_after == 'xxo' or S_after == 'xox':
                    ans = 'FAILURE'
                else:
                    ans = 'SUCCESS'    
            if S_before == 'xoo':
                if S_after == 'xoo' or S_after == 'oxo':
                    ans = 'FAILURE'
                else:
                    ans = 'SUCCESS'
            if S_before == 'oxo':
                if S_after == 'xoo' or S_after == 'oox':
                    ans = 'FAILURE'
                else:
                    ans = 'SUCCESS'            
            if S_before == 'oox':
                if S_after == 'oox' or S_after == 'oxo':
                    ans = 'FAILURE'
                else:
                    ans = 'SUCCESS'  
print(ans)