题目明确表示了从前往后依次运算，所以其实从后往前看，始终是(ABC)+/-/*//D的形式，所以dfs采用倒序遍历的方式，最末尾的数直接影响前面几个数应该的输出值，直接倒序打印遍历，即可找到对应的计算流程了 
a = input().split(' ')
dictory = {'J':11,'Q':12,'K':13,'A':1}
def dfs(wait,target,out):
    if len(wait) == 1:
        if wait[0] in dictory:
            c = dictory[wait[0]]
        else:
            c = int(wait[0])
        if target == c:
            L.append(wait[0]+out)
    else:
        for i in range(len(wait)):
            w = wait[:i]+wait[i+1::]
            if wait[i] in dictory:
                c = dictory[wait[i]]
            else:
                c = int(wait[i])
            dfs(w,target-c,'+'+wait[i]+out)
            dfs(w,target+c,'-'+wait[i]+out)
            dfs(w,target*c,'/'+wait[i]+out)
            dfs(w,target/c,'*'+wait[i]+out)

L = []
if 'joker' in a or 'JOKER' in a:
    print('ERROR')
else:
    dfs(a,24,'')
    if not L:
        print('NONE')
    else:
        print(L[0])