A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".

We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").

The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.

A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.

Notice, that any year between 2013 and 2015 is not a leap year.

Input

The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.

Output

In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.

Examples

Input

777-444---21-12-2013-12-2013-12-2013---444-777


Output

13-12-2013
s = input()
months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
n = len(s)
d = {}
max_k = 0
max_t = ''
for i in range(n - 9):
    if s[i + 2] == '-' and s[i + 5] == '-':
        if s[i + 3: i + 5].isdigit() and 1 <= int(s[i + 3: i + 5]) <= 12:
            m = int(s[i + 3: i + 5])
            if s[i: i + 2].isdigit() and 1 <= int(s[i: i + 2]) <= months[m - 1]:
                if s[i + 6: i + 10].isdigit() and 2013 <= int(s[i + 6: i + 10]) <= 2015:
                    t = s[i: i + 10]
                    if d.get(t):
                        d[t] += 1
                    else:
                        d[t] = 1
                    if d[t] >= max_k:
                        max_k = d[t]
                        max_t = t
print(max_t)
