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[100006], wc[13];
bool ok[10];
long D[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
map<string, long> MAP;
long anss;
string ans, now;
int main() {
  std::ios::sync_with_stdio(false);
  long i, j, d, m, y;
  cin >> s;
  ok[2] = ok[5] = 1;
  for (i = 0; s[i + 9] != '\0'; ++i) {
    for (j = 0; j < 10; ++j)
      if ((s[j + i] == '-') != ok[j]) break;
    if (j == 10) {
      for (j = 0; j < 10; ++j) wc[j] = s[j + i];
      wc[10] = '\0';
      now = wc;
      d = (now[0] - '0') * 10 + (now[1] - '0');
      m = (now[3] - '0') * 10 + (now[4] - '0');
      y = (now[6] - '0') * 1000 + (now[7] - '0') * 100 + (now[8] - '0') * 10 +
          (now[9] - '0');
      if (m > 12 || m == 0 || y < 2013 || y > 2015) continue;
      if (d > D[m] || d == 0) continue;
      if (++MAP[now] > anss) {
        anss = MAP[now];
        ans = now;
      }
    }
  }
  cout << ans << endl;
  return 0;
}
