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
#include <bits/stdc++.h>
using namespace std;
int main() {
  string s, ans;
  char t[11];
  cin >> s;
  int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  int len = 0, day = 0, mon = 0, year = 0, flag = 0, cnt, max = 0;
  len = s.length();
  for (year = 2013; year <= 2015; year++) {
    for (mon = 1; mon <= 12; mon++) {
      for (day = 1; day <= m[mon]; day++) {
        sprintf(t, "%02d-%02d-%04d", day, mon, year);
        cnt = 0;
        for (int i = 0; i <= len - 10; i++) {
          flag = 0;
          for (int j = 0; j < 10; j++) {
            if (t[j] != s[j + i]) {
              flag = 1;
              break;
            }
          }
          if (flag == 0) {
            cnt++;
          }
        }
        if (cnt > max) {
          max = cnt;
          ans = t;
        }
      }
    }
  }
  cout << ans;
  return 0;
}
