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 Day[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int maxx = -1;
int main() {
  string s, ans;
  cin >> s;
  map<string, int> mp;
  for (int i = 0; i < (int)s.size() - 9; i++) {
    string t = s.substr(i, 10);
    int year =
        (t[9] - 48) + (t[8] - 48) * 10 + (t[7] - 48) * 100 + (t[6] - 48) * 1000;
    int month = (t[4] - 48) + (t[3] - 48) * 10;
    int day = (t[1] - 48) + (t[0] - 48) * 10;
    if (year > 2015 || year < 2013 || month > 12 || month < 1 ||
        day > Day[month] || day < 1 || count(t.begin(), t.end(), '-') != 2) {
      continue;
    }
    mp[t]++;
    if (mp[t] > maxx) {
      maxx = mp[t];
      ans = t;
    }
  }
  cout << ans << endl;
  return 0;
}
