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;
long long gcd(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
int main() {
  long long mon[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  string s, ans;
  cin >> s;
  long long ma = 0;
  for (long long i = 2013; i < 2016; i++) {
    for (long long j = 1; j < 13; j++) {
      for (long long k = 1; k <= mon[j - 1]; k++) {
        long long c = 0;
        string t;
        if (k < 10) t = "0";
        t = t + to_string(k) + "-";
        if (j < 10) t = t + "0";
        t = t + to_string(j) + "-";
        t = t + to_string(i);
        size_t f = 0;
        while ((f = s.find(t, f)) != -1) {
          c++;
          f++;
        }
        if (c > ma) {
          ma = c;
          ans = t;
        }
      }
    }
  }
  cout << ans;
  return 0;
}
