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;
string a, b;
unordered_map<string, int> mm;
int day[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool check(string a) {
  for (int i = 0; i < a.size(); i++) {
    if (i == 2 || i == 5) {
      if (a[i] != '-') {
        return false;
      }
    } else {
      if (a[i] == '-') {
        return false;
      }
    }
  }
  int nn = 10 * (a[3] - '0') + a[4] - '0';
  if (day[nn] < 10 * (a[0] - '0') + a[1] - '0' ||
      10 * (a[0] - '0') + a[1] - '0' == 0 || nn == 0) {
    return false;
  }
  nn =
      1000 * (a[6] - '0') + 100 * (a[7] - '0') + 10 * (a[8] - '0') + a[9] - '0';
  if (nn < 2013 || nn > 2015) {
    return false;
  }
  return true;
}
string ans;
int cc = 0;
int main() {
  cin >> a;
  for (int i = 0; i <= a.size() - 10; i++) {
    b = a.substr(i, 10);
    if (check(b)) {
      mm[b]++;
      if (mm[b] > cc) {
        cc = mm[b];
        ans = b;
      }
    }
    b.clear();
  }
  cout << ans;
  return 0;
}
