题解 | #MP3光标位置# 少于四首歌的情况好容易漏啊

n = int(input())
order = list(input())

#分成两部分
#计算当前歌曲 这个相当于一个取余循环  记录总的操作数取余就行了
#计算当前光标是在第几行[1,2,3,4]
#初始是1 D [1,2,3]->[2,3,4]  4位置翻页不翻页根据当前歌曲num == n讨论一下

num_p = 0
# index  0 1 2 3 4  ... n-1  -1=> n-1
# value  1 2 3 4 5  ... n
# num = (num_p+n) % n + 1
local_p = 1
#1 2 3 4

for od in order:
    if od == 'D':
        num = (num_p+n) % n + 1

        num_p += 1

        if local_p in [1, 2, 3]:
            local_p = local_p+1
        elif local_p == 4:
            # 到底翻页
            local_p = 1 if num == n else 4

    else:
        num = (num_p+n) % n + 1

        num_p -= 1

        if local_p in [2, 3, 4]:
            local_p = local_p-1
        elif local_p == 1:
            # 到顶翻页
            local_p = 4 if num == 1 else 1



num = (num_p+n) % n + 1
left = local_p - 1
right = 4 - local_p
list_out = [i for i in range(num-left, num+right+1)] if n>=4 else [i+1 for i in range(n)]


print(" ".join([str(i) for i in list_out]))
print(num)