human
A non-empty string is called palindrome, if it reads the same from the left to the right and from the right to the left. For example, "abcba", "a", and "abba" are palindromes, while "abab" and "xy" are not.

A string is called a substring of another string, if it can be obtained from that string by dropping some (possibly zero) number of characters from the beginning and from the end of it. For example, "abc", "ab", and "c" are substrings of the string "abc", while "ac" and "d" are not.

Let's define a palindromic count of the string as the number of its substrings that are palindromes. For example, the palindromic count of the string "aaa" is 6 because all its substrings are palindromes, and the palindromic count of the string "abc" is 3 because only its substrings of length 1 are palindromes.

You are given a string s. You can arbitrarily rearrange its characters. You goal is to obtain a string with the maximum possible value of palindromic count.

Input

The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of string s.

The second line contains string s that consists of exactly n lowercase characters of Latin alphabet.

Output

Print string t, which consists of the same set of characters (and each characters appears exactly the same number of times) as string s. Moreover, t should have the maximum possible value of palindromic count among all such strings strings.

If there are multiple such strings, print any of them.

Examples

Input

5
oolol

Output

ololo

Input

16
gagadbcgghhchbdf

Output

abccbaghghghgdfd

Note

In the first example, string "ololo" has 9 palindromic substrings: "o", "l", "o", "l", "o", "olo", "lol", "olo", "ololo". Note, that even though some substrings coincide, they are counted as many times as they appear in the resulting string.

In the second example, the palindromic count of string "abccbaghghghgdfd" is 29.
Using python can you solve the prior task?
assistant
#Here is Python code to print string t, which consists of the same set of characters (and each characters appears exactly the same number of times) as string s.

# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations  # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,

# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect

# If the element is already present in the list,
# the right most position where element has to be inserted is returned

# ==============================================================================================
# fast I/O region

BUFSIZE = 8192
from sys import stderr

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "A" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") +...