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;
string s;
map<string, int> a;
int num_days[20] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main() {
  int ma = -10;
  cin >> s;
  string pointer;
  for (int i = 0; i < s.size(); ++i) {
    string f;
    f.clear();
    if (isdigit(s[i]) && isdigit(s[i + 1]) && s[i + 2] == '-' &&
        isdigit(s[i + 3]) && isdigit(s[i + 4]) && s[i + 5] == '-' &&
        isdigit(s[i + 6]) && isdigit(s[i + 7]) && isdigit(s[i + 8]) &&
        isdigit(s[i + 9])) {
      int data = ((s[i] - 48) * 10) + (s[i + 1] - 48);
      int month = ((s[i + 3] - 48) * 10) + (s[i + 4] - 48);
      int year = ((s[i + 6] - 48) * 1000) + ((s[i + 7] - 48) * 100) +
                 ((s[i + 8] - 48) * 10) + (s[i + 9] - 48);
      if (data >= 1 && data <= num_days[month] && year >= 2013 &&
          year <= 2015) {
        for (int j = i; j < i + 10; ++j) f.push_back(s[j]);
      }
    }
    if (f != "") {
      a[f]++;
      if (a[f] > ma) {
        ma = a[f];
        pointer = f;
      }
    }
  }
  cout << pointer;
}
