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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int INF = 1 << 30;
const int base = 1000 * 1000 * 1000;
const int N = 1100000;
const int DAYS[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
char s[110000] = {};
map<string, int> M;
int len = 0;
void Solve() {
  char date[11] = "01-01-2013";
  for (int i = 3; i < 6; i++)
    for (int j = 1; j < 13; j++)
      for (int day = 1; day <= DAYS[j - 1]; day++) {
        date[0] = day / 10 + '0';
        date[1] = day % 10 + '0';
        date[3] = j / 10 + '0';
        date[4] = j % 10 + '0';
        date[9] = i + '0';
        int Count = 0;
        for (int k = 0; k < len - 9; k++)
          if (strncmp(&s[k], date, 10) == 0) Count++;
        M[date] = Count;
      }
}
int main() {
  gets(s);
  len = strlen(s);
  Solve();
  string answ;
  int Max = -1;
  for (map<string, int>::iterator it = M.begin(); it != M.end(); it++)
    if (it->second > Max) {
      answ = it->first;
      Max = it->second;
    }
  for (int i = 0; i < answ.size(); i++) putchar(answ[i]);
  putchar('\n');
  return 0;
}
