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[100010];
int mon[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int anum;
map<string, int> ms;
bool isnum(char a) { return a >= '0' && a <= '9'; }
int main() {
  while (~scanf("%s", s)) {
    ms.clear();
    anum = 0;
    int tmp = 0;
    string tt;
    int max = 0;
    for (int i = 0; s[i + 9] != 0; i++) {
      if (isnum(s[i]) && isnum(s[i + 1]) && !isnum(s[i + 2]) &&
          isnum(s[i + 3]) && isnum(s[i + 4]) && !isnum(s[i + 5]) &&
          isnum(s[i + 6]) && isnum(s[i + 7]) && isnum(s[i + 8]) &&
          isnum(s[i + 9])) {
        int a0 = atoi(s + i), a1 = atoi(s + i + 3), a2;
        char ttt = s[i + 10];
        s[i + 10] = 0;
        a2 = atoi(s + i + 6);
        s[i + 10] = ttt;
        if (a2 <= 2015 && a2 >= 2013 && a1 <= 12 && a1 >= 1 && a0 > 0 &&
            a0 <= mon[a1]) {
          ms[string(s + i, s + i + 10)]++;
          if (ms[string(s + i, s + i + 10)] > max) {
            tt = string(s + i, s + i + 10);
            max = ms[tt];
          }
        }
      }
    }
    printf("%s", tt.c_str());
    puts("");
  }
}
