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;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
string ans, s;
int len, L;
inline void check(int y, int m, int d) {
  string res = "";
  for (int i = 1; i <= 4; ++i) {
    res += char(y % 10 + '0');
    y /= 10;
  }
  res += '-';
  for (int i = 1; i <= 2; ++i) {
    res += char(m % 10 + '0');
    m /= 10;
  }
  res += '-';
  for (int i = 1; i <= 2; ++i) {
    res += char(d % 10 + '0');
    d /= 10;
  }
  reverse(res.begin(), res.end());
  if (s.find(res) == string::npos) return;
  int book = 0;
  for (int i = 0; i < L; ++i) {
    if (s.substr(i, 10) == res) {
      book++;
      if (book > len) {
        len = book;
        ans = res;
      }
    }
  }
}
int main() {
  cin >> s;
  L = s.size();
  for (int y = 2013; y <= 2015; ++y) {
    for (int m = 1; m <= 12; ++m) {
      for (int d = 1; d <= mon[m]; ++d) {
        check(y, m, d);
      }
    }
  }
  cout << ans << endl;
  return 0;
}
