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>
const int S[] = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};
int DtCvt(char *z, bool *F) {
  static const bool H[10] = {0, 0, 1, 0, 0, 1};
  if (10 + H != std::mismatch(H, 10 + H, F).first) return -1;
  static const int M[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  int y, m, d;
  sscanf(z, "%d", &d), sscanf(3 + z, "%d", &m), sscanf(6 + z, "%4d", &y);
  return 0 > (y -= 2013) || 2 < y || !m || 11 < --m || !d || M[m] < d--
             ? -1
             : d + S[m] + 365 * y;
}
void Output(int s) {
  int y = 2013 + s / 365, m = std::upper_bound(1 + S, 12 + S, s %= 365) - S,
      d = 1 + s - S[m - 1];
  printf("%02i-%02i-%i\n", d, m, y);
}
int main(void) {
  bool F[100000];
  char z[100001];
  for (long n, m, S[1095]; gets(z); Output(std::max_element(S, 1095 + S) - S)) {
    for (n = -1; z[++n]; F[n] = '-' == z[n])
      ;
    for (memset(S, 0, sizeof(S)), n -= 9; n--;)
      if (~(m = DtCvt(n + z, n + F))) ++S[m];
  }
  return 0;
}
