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;
char s[100005];
int day[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
map<string, int> my;
void check(char* s) {
  if (((s[0]) >= '0' && (s[0]) <= '9') && ((s[1]) >= '0' && (s[1]) <= '9') &&
      ((s[3]) >= '0' && (s[3]) <= '9') && ((s[4]) >= '0' && (s[4]) <= '9') &&
      ((s[6]) >= '0' && (s[6]) <= '9') && ((s[7]) >= '0' && (s[7]) <= '9') &&
      ((s[8]) >= '0' && (s[8]) <= '9') && ((s[9]) >= '0' && (s[9]) <= '9') &&
      s[2] == '-' && s[5] == '-') {
    int d = atoi(s);
    int m = atoi(s + 3);
    int y = atoi(string(s + 6, s + 10).c_str());
    if (m <= 12 && d > 0 && d <= day[m - 1] && y >= 2013 && y <= 2015) {
      ++my[string(s, s + 10)];
    }
  }
}
int main() {
  gets(s);
  int n = strlen(s);
  for (int i = 0; i + 10 <= n; ++i) check(s + i);
  string res("");
  int max = -1;
  for (typeof((my).begin()) it = (my).begin(), _e = (my).end(); it != _e; ++it)
    if (it->second > max) {
      max = it->second;
      res = it->first;
    }
  cout << res << endl;
  return 0;
}
