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 months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
string s;
int main() {
  while (getline(cin, s)) {
    int n = (int)s.size();
    vector<string> ans;
    for (int i = 0; i <= n - 10; i++) {
      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 day = (s[i] - '0') * 10 + s[i + 1] - '0';
        int month = (s[i + 3] - '0') * 10 + s[i + 4] - '0';
        int year = (s[i + 6] - '0') * 1000 + (s[i + 7] - '0') * 100 +
                   (s[i + 8] - '0') * 10 + s[i + 9] - '0';
        if (year >= 2013 && year <= 2015 && month >= 1 && month <= 12 &&
            day >= 1 && day <= months[month])
          ans.push_back(s.substr(i, 10));
      }
    }
    sort((ans).begin(), (ans).end());
    string best = "";
    int cnt = 0;
    int i = 0;
    while (i < (int)ans.size()) {
      int c = 1;
      while (i < (int)ans.size() - 1 && ans[i] == ans[i + 1]) ++i, ++c;
      if (cnt < c) best = ans[i], cnt = c;
      ++i;
    }
    printf("%s\n", best.c_str());
  }
  return 0;
}
