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<string, int> f;
int k[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool invalid(char a) { return a == '-'; }
void calc(string s) {
  if (s[2] != '-' || s[5] != '-') return;
  for (int i = 0; i < s.length(); i++)
    if (i == 2 || i == 5)
      continue;
    else if (invalid(s[i]))
      return;
  string d = s.substr(0, 2), m = s.substr(3, 2), y = s.substr(6, 4);
  int year = atoi(y.c_str()), month = atoi(m.c_str()), day = atoi(d.c_str());
  if (year < 2013 || year > 2015 || month < 1 || month > 12 || day < 1 ||
      day > k[month])
    return;
  f[s]++;
}
int main() {
  int maxi = -1;
  string s, MX = "";
  cin >> s;
  for (int i = 0; i < s.size() - 9; i++) calc(s.substr(i, 10));
  for (map<string, int>::iterator it = f.begin(); it != f.end(); it++)
    if (it->second > maxi) maxi = it->second, MX = it->first;
  cout << MX << endl;
  return 0;
}
