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;
map<int, int> m;
map<int, int>::iterator it;
int mesec(int m) {
  if (m == 1)
    return 31;
  else if (m == 2)
    return 28;
  else if (m == 3)
    return 31;
  else if (m == 4)
    return 30;
  else if (m == 5)
    return 31;
  else if (m == 6)
    return 30;
  else if (m == 7)
    return 31;
  else if (m == 8)
    return 31;
  else if (m == 9)
    return 30;
  else if (m == 10)
    return 31;
  else if (m == 11)
    return 30;
  else if (m == 12)
    return 31;
  else
    return -1;
}
int main() {
  char a[100011];
  cin.get(a, 100011);
  long long int q1, q2, q3, q4, q5, mx = 0;
  for (int i = 2; i < strlen(a); i++) {
    if (a[i] == '-') {
      if (a[i - 1] >= '0' && a[i - 1] <= '9')
        if (a[i - 2] >= '0' && a[i - 2] <= '9')
          if (a[i + 1] >= '0' && a[i + 1] <= '9')
            if (a[i + 2] >= '0' && a[i + 2] <= '9')
              if (a[i + 3] == '-' && a[i + 3] == '-')
                if (a[i + 4] >= '0' && a[i + 4] <= '9')
                  if (a[i + 5] >= '0' && a[i + 5] <= '9')
                    if (a[i + 6] >= '0' && a[i + 6] <= '9')
                      if (a[i + 7] >= '0' && a[i + 7] <= '9') {
                        q1 = (a[i - 2] - '0') * 10 + (a[i - 1] - '0');
                        q2 = (a[i + 1] - '0') * 10 + (a[i + 2] - '0');
                        q3 = (a[i + 4] - '0') * 1000 + (a[i + 5] - '0') * 100 +
                             (a[i + 6] - '0') * 10 + (a[i + 7] - '0');
                        if (q1 >= 1 && q1 <= mesec(q2))
                          if (q3 >= 2013 && q3 <= 2015) {
                            q4 = q1 * 1000000 + q2 * 10000 + q3;
                            m[q4]++;
                            if (m[q4] > mx) {
                              mx = m[q4];
                              q5 = q4;
                            }
                          }
                      }
    }
  }
  if (q5 / 1000000 < 10) cout << 0;
  cout << q5 / 1000000 << "-";
  if (q5 / 10000 % 100 < 10) cout << 0;
  cout << q5 / 10000 % 100 << "-";
  cout << q5 % 10000 << endl;
  return 0;
}
