It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.

Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.

Input

The single line contains integer y (1000 ≤ y ≤ 9000) — the year number.

Output

Print a single integer — the minimum year number that is strictly larger than y and all it's digits are distinct. It is guaranteed that the answer exists.

Examples

Input

1987


Output

2013


Input

2013


Output

2014
import re
def checkYear(fnumber):
	zero = len(re.findall("0" , fnumber)) 
	one = len(re.findall("1" , fnumber))
	two = len(re.findall("2" , fnumber))
	three = len(re.findall("3" , fnumber))
	four = len(re.findall("4" , fnumber))
	five = len(re.findall("5" , fnumber))
	six = len(re.findall("6" , fnumber))
	seven = len(re.findall("7" , fnumber))
	eight = len(re.findall("8" , fnumber))
	nine = len(re.findall("9" , fnumber))
	if zero > 1 or one > 1 or two > 1 or one > 1 or two > 1 or three > 1 or four > 1 or five > 1 or six > 1 or seven > 1 or eight > 1 or nine > 1:
		return False
	else:
		return True

year = int(input())
found = False
i = year + 1
while not found : 
	if checkYear(str(i)) :
		found = True
		answer = i
	else : 
		i += 1
print(answer)