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;
string s;
int arr[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
pair<int, int> f(int i) {
  if (i - 6 >= 0 && s[i - 4] == '-' && s[i - 1] == '-' && s[i - 3] >= '0' &&
      s[i - 3] <= '9' && s[i - 2] >= '0' && s[i - 2] <= '9' &&
      s[i - 6] >= '0' && s[i - 6] <= '9' && s[i - 5] >= '0' &&
      s[i - 5] <= '9') {
    int x = (s[i - 3] - '0') * 10 + (s[i - 2] - '0');
    if (x > 0 && x <= 12) {
      int d = (s[i - 6] - '0') * 10 + (s[i - 5] - '0');
      if (d > 0 && d <= arr[x - 1]) return make_pair(d, x);
    }
  }
  return make_pair(-1, -1);
}
int main() {
  map<string, int> M;
  cin >> s;
  int v;
  int d = 0;
  string ans = "";
  for (int i = 0; i < s.size(); i++) {
    if (i + 3 < s.size() && s[i] == '2' && s[i + 1] == '0' && s[i + 2] == '1' &&
        (s[i + 3] >= '3' && s[i + 3] <= '5')) {
      pair<int, int> p = f(i);
      if (p.first != -1) {
        string a = "";
        for (int j = i - 6; j < i + 4; j++) a += s[j];
        if (M.count(a) == 0) {
          M[a] = 1;
          if (1 > d) {
            d = 1;
            ans = a;
          }
        } else {
          M[a]++;
          int t = M[a];
          if (t > d) {
            d = t;
            ans = a;
          }
        }
      }
    }
  }
  cout << ans;
  return 0;
}
