def max_OU(n, m, s):
    left = 0
    right = 0
    max_ou = 0
    current_ou = 0
    k_used = False

    while right < n:
        if s[right] == 'O':
            o_count = 1
            # 计算窗口内"O"的数量
            for i in range(right + 1, n):
                if s[i] == 'O':
                    o_count += 1
                elif s[i] == 'K':
                    break
                elif s[i] == 'U':
                    break
            u_count = 0
            # 计算窗口内"U"的数量
            for i in range(right + 1, n):
                if s[i] == 'U':
                    u_count += 1
                elif s[i] == 'K':
                    if not k_used:
                        u_count += o_count
                        k_used = True
                        break
                elif s[i] == 'O':
                    break
            current_ou += min(o_count, u_count)
            max_ou = max(max_ou, current_ou)
        elif s[right] == 'U':
            if current_ou > 0:
                current_ou -= 1
        elif s[right] == 'K' and not k_used:
            k_used = True
            # 尝试删除"K"，看是否能增加"OU"对的数量
            tmp_right = right
            while tmp_right < n and s[tmp_right] != 'U':
                tmp_right += 1
            if tmp_right < n:
                current_ou += 1
                max_ou = max(max_ou, current_ou)

        right += 1
        if current_ou > 0 and left < right and s[left] == 'O':
            current_ou -= 1
            left += 1

    return max_ou

# 读取输入
n, k = map(int, input().split())
s = input()

# 计算并输出结果
print(max_OU(n, k, s))