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;
int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
map<string, int> m;
bool isok(char *ss) {
  int i = 0, cnt = 0;
  for ((i) = 0; (i) < (10); (i)++) {
    if (!isdigit(*(ss + i))) cnt++;
  }
  if (cnt == 2 && *(ss + 2) == '-' && *(ss + 5) == '-') return true;
  return false;
}
bool isvalid(char *ss) {
  int d, m, y;
  sscanf(ss, "%2d%*c%2d%*c%4d", &d, &m, &y);
  return (y >= 2013 && y <= 2015) && (m > 0 && m <= 12) &&
         (d >= 1 && d <= days[m - 1]);
}
char s[100005];
int main() {
  scanf("%s", s);
  int n = strlen(s), i, mx = 0, j;
  string ans;
  for ((i) = 0; (i) < (n - 9); (i)++) {
    if (isok(s + i) && isvalid(s + i)) {
      string a = "";
      for ((j) = 0; (j) < (10); (j)++) a.push_back(s[j + i]);
      if (m.find(a) == m.end())
        m[a] = 1;
      else
        m[a]++;
      if (m[a] > mx) {
        mx = m[a];
        ans = a;
      }
    }
  }
  cout << ans << endl;
  return 0;
}
