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>
char a[100000 + 2];
int vis[32][13][6];
int save[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
void solve() {
  memset(vis, 0, sizeof(vis));
  int len = strlen(a);
  int max = 0, d, m, y;
  for (int i = 9; i < len; i++) {
    if ((a[i] == '3' || a[i] == '4' || a[i] == '5') && a[i - 1] == '1' &&
        a[i - 2] == '0' && a[i - 3] == '2') {
      if (a[i - 4] != '-' || a[i - 7] != '-' || a[i - 5] == '-' ||
          a[i - 6] == '-' || a[i - 8] == '-' || a[i - 9] == '-')
        continue;
      int dd = (a[i - 9] - '0') * 10 + (a[i - 8] - '0');
      int mm = (a[i - 6] - '0') * 10 + (a[i - 5] - '0');
      if (mm <= 12 && mm > 0 && dd > 0 && dd <= save[mm - 1]) {
        int k = ++vis[dd][mm][a[i] - '0'];
        if (k > max) {
          max = k;
          d = dd;
          m = mm;
          y = 2010 + a[i] - '0';
        }
      }
    }
  }
  if (d < 10) printf("0");
  printf("%d-", d);
  if (m < 10) printf("0");
  printf("%d-%d\n", m, y);
}
int main() {
  while (gets(a)) {
    solve();
  }
  return 0;
}
