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 str[123456];
char it[123];
char day[13][5] = {"31", "28", "31", "30", "31", "30",
                   "31", "31", "30", "31", "30", "31"};
map<string, int> m;
map<string, int>::iterator ut;
int main() {
  scanf("%s", str);
  int len = strlen(str);
  m.clear();
  for (int i = 0; i < len - 3; i++) {
    if (str[i] == '-' && str[i + 3] == '-' && str[i + 4] == '2' &&
        str[i + 5] == '0' && str[i + 6] == '1') {
      int flag = 0;
      for (int j = 0; j < 10; j++) {
        it[j] = str[i + j - 2];
      }
      it[10] = '\0';
      if (it[9] < '3' || it[9] > '5') flag = 1;
      if (it[3] < '0' || it[3] > '9' || it[3] > '1') flag = 1;
      if (it[3] == '0') {
        if (it[4] < '0' || it[4] > '9' || it[4] == '0') flag = 1;
      }
      if (it[3] == '1') {
        if (it[4] < '0' || it[4] > '9' || it[4] > '2') flag = 1;
      }
      if (it[0] < '0' || it[0] > '9') flag = 1;
      if (it[0] == '0') {
        if (it[1] <= '0' || it[1] > '9') flag = 1;
      } else {
        if (it[1] < '0' || it[1] > '9') flag = 1;
      }
      int u = (it[3] - '0') * 10 + (it[4] - '0');
      char t[3];
      t[0] = it[0], t[1] = it[1];
      t[2] = '\0';
      if (strcmp(t, day[u - 1]) > 0) flag = 1;
      if (!flag) {
        string t = "";
        for (int j = 0; j < 10; j++) t += it[j];
        m[t]++;
      }
    }
  }
  string ans;
  int u = -1;
  for (ut = m.begin(); ut != m.end(); ut++) {
    if (ut->second > u) {
      u = ut->second;
      ans = ut->first;
    }
  }
  cout << ans << endl;
  return 0;
}
