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 Days(int Month) {
  if (Month == 2) return 28;
  if (Month == 4 || Month == 6 || Month == 9 || Month == 11) return 30;
  return 31;
}
int Calc(string S) {
  if (S[2] != '-' || S[5] != '-') return -1;
  for (int i = 0; i < 10; i++) {
    if (i == 2 || i == 5) continue;
    if (S[i] < '0' || S[i] > '9') return -1;
  }
  int Day = (S[0] - 48) * 10 + (S[1] - 48);
  int Month = (S[3] - 48) * 10 + (S[4] - 48);
  int Year =
      (S[6] - 48) * 1000 + (S[7] - 48) * 100 + (S[8] - 48) * 10 + (S[9] - 48);
  if (Year < 2013 || Year > 2015) return -1;
  if (Month < 1 || Month > 12) return -1;
  if (Day < 1 || Day > Days(Month)) return -1;
  return (Year - 2013) * 10000 + Month * 100 + Day;
}
void Output(int X) {
  putchar(X / 10 + 48);
  putchar(X % 10 + 48);
}
int main() {
  string S;
  cin >> S;
  static int Count[30000];
  memset(Count, 0, sizeof(Count));
  for (int i = 0; i + 10 <= S.size(); i++) {
    string T = "";
    for (int j = 0; j < 10; j++) T += S[i + j];
    int Temp = Calc(T);
    if (Temp != -1) Count[Temp]++;
  }
  int Ans = 0;
  for (int i = 1; i < 30000; i++)
    if (Count[i] > Count[Ans]) Ans = i;
  Output(Ans % 100);
  putchar('-');
  Output(Ans / 100 % 100);
  putchar('-');
  printf("%d\n", Ans / 10000 + 2013);
  return 0;
}
