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;
int cnt[40][20][10];
int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main() {
  cin >> a;
  int n = a.size(), mx = 0, ansd = 0, ansm = 0, ansy = 0;
  for (int i = 0; i + 9 < n; i++) {
    if (a[i] == '-' || a[i + 1] == '-' || a[i + 2] != '-' || a[i + 3] == '-' ||
        a[i + 4] == '-' || a[i + 5] != '-' || a[i + 6] != '2' ||
        a[i + 7] != '0' || a[i + 8] != '1' || a[i + 9] == '-')
      continue;
    int dd = (a[i] - 48) * 10 + a[i + 1] - 48;
    int mm = (a[i + 3] - 48) * 10 + a[i + 4] - 48;
    int yy = a[i + 9] - 48;
    if (3 <= yy && yy <= 5 && 1 <= mm && mm <= 12 && 1 <= dd && dd <= m[mm]) {
      cnt[dd][mm][yy]++;
      if (cnt[dd][mm][yy] > mx) {
        mx = cnt[dd][mm][yy];
        ansd = dd;
        ansm = mm;
        ansy = yy;
      }
    }
  }
  printf("%d%d-%d%d-%d\n", ansd / 10, ansd % 10, ansm / 10, ansm % 10,
         2010 + ansy);
  return 0;
}
