User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are *n* pages numbered by integers from 1 to *n*. Assume that somebody is on the *p*-th page now. The navigation will look like this:

When someone clicks the button "&lt;&lt;" he is redirected to page 1, and when someone clicks the button "&gt;&gt;" he is redirected to page *n*. Of course if someone clicks on a number, he is redirected to the corresponding page.

There are some conditions in the navigation:
 -  If page 1 is in the navigation, the button "&lt;&lt;" must not be printed. -  If page *n* is in the navigation, the button "&gt;&gt;" must not be printed. -  If the page number is smaller than 1 or greater than *n*, it must not be printed. 
You can see some examples of the navigations. Make a program that prints the navigation.
The first and the only line contains three integers *n*, *p*, *k* (3 ≤ *n* ≤ 100; 1 ≤ *p* ≤ *n*; 1 ≤ *k* ≤ *n*)
nums = input().split(' ')
n, p, k = [int(i) for i in nums]

front = last = ''
if p - k > 1:
    front = '<< '
if p + k < n:
    last = ' >>'

f, l = [], []
for i in range(max(1, p - k), p):
    f.append(str(i))
for i in range(p + 1, 1 + min(n, p + k)):
    l.append(str(i))
res = front + ' '.join(f + ['(' + str(p) + ')'] + l) + last
print(res)
# class Solution(object):
#     def countSquares(self, matrix):
#         """
#         :type matrix: List[List[int]]
#         :rtype: int
#         """
#         res = 0
#         row, col = len(matrix), len(matrix[0])
#         for i in range(1, row):
#             for j in range(1, col):
#                 if matrix[i][j] != 0:
#                     matrix[i][j] += min(matrix[i - 1][j - 1], matrix[i][j - 1], matrix[i - 1][j])
#                     # if matrix[i - 1][j - 1] and matrix[i][j - 1] and matrix[i - 1][j]:
#                     #     res += 1
#         for i in range(row):
#             for j in range(col):
#                 if matrix[i][j]:
#                     res += matrix[i][j]
#         print(matrix)
#         return res
        
            

# matrix = [
#   [0,1,1,1],
#   [1,1,1,1],
#   [0,1,1,1]
# ]

# a = Solution().countSquares(matrix)
# print(a)