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 main() {
  vector<string> subs;
  int l;
  string a, b;
  cin >> a;
  l = a.size();
  for (int i = 0; i < l - 9; ++i) {
    b = a.substr(i, 10);
    if ((b[2] == '-' && b[5] == '-') &&
        ((((b[0] - '0') * 10 + b[1] - '0') <= 28) &&
         (((b[0] - '0') * 10 + b[1] - '0') > 0)) &&
        ((((b[3] - '0') * 10 + b[4] - '0') <= 12) &&
         (((b[3] - '0') * 10 + b[4] - '0') > 0)) &&
        ((b[6] == '2') && (b[7] == '0') && (b[8] == '1') &&
         ((b[9] - '0') >= 3) && ((b[9] - '0') <= 5)) &&
        (((b[0] >= '0') && (b[0] <= '9')) && ((b[1] >= '0') && (b[1] <= '9')) &&
         ((b[3] >= '0') && (b[3] <= '9')) &&
         ((b[4] >= '0') && (b[4] <= '9')))) {
      subs.push_back(b);
    } else if ((((b[3] - '0') * 10 + b[4] - '0') != 2) &&
               (b[2] == '-' && b[5] == '-') &&
               ((((b[0] - '0') * 10 + b[1] - '0') > 28) &&
                (((b[0] - '0') * 10 + b[1] - '0') <= 30)) &&
               ((((b[3] - '0') * 10 + b[4] - '0') <= 12) &&
                (((b[3] - '0') * 10 + b[4] - '0') > 0)) &&
               ((b[6] == '2') && (b[7] == '0') && (b[8] == '1') &&
                ((b[9] - '0') >= 3) && ((b[9] - '0') <= 5)) &&
               (((b[0] >= '0') && (b[0] <= '9')) &&
                ((b[1] >= '0') && (b[1] <= '9')) &&
                ((b[3] >= '0') && (b[3] <= '9')) &&
                ((b[4] >= '0') && (b[4] <= '9')))) {
      subs.push_back(b);
    } else if ((((b[3] - '0') * 10 + b[4] - '0') != 11) &&
               (((b[3] - '0') * 10 + b[4] - '0') != 9) &&
               (((b[3] - '0') * 10 + b[4] - '0') != 6) &&
               (((b[3] - '0') * 10 + b[4] - '0') != 4) &&
               (((b[3] - '0') * 10 + b[4] - '0') != 2) &&
               (b[2] == '-' && b[5] == '-') &&
               ((((b[0] - '0') * 10 + b[1] - '0') == 31)) &&
               ((((b[3] - '0') * 10 + b[4] - '0') <= 12) &&
                (((b[3] - '0') * 10 + b[4] - '0') > 0)) &&
               ((b[6] == '2') && (b[7] == '0') && (b[8] == '1') &&
                ((b[9] - '0') >= 3) && ((b[9] - '0') <= 5)) &&
               (((b[0] >= '0') && (b[0] <= '9')) &&
                ((b[1] >= '0') && (b[1] <= '9')) &&
                ((b[3] >= '0') && (b[3] <= '9')) &&
                ((b[4] >= '0') && (b[4] <= '9')))) {
      subs.push_back(b);
    }
  }
  l = subs.size();
  sort(subs.begin(), subs.end());
  subs.push_back("");
  int cmax = -1;
  int ccur = 0;
  b = "";
  for (int i = 0; i < l; ++i) {
    if (subs[i] == subs[i + 1]) {
      ccur++;
    } else {
      if (ccur > cmax) {
        cmax = ccur;
        b = subs[i];
      }
      ccur = 0;
    }
  }
  cout << b << endl;
  return 0;
}
