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;
const int inf = (int)1e9;
const double EPS = 1e-9, INF = 1e12;
string in;
int day(int y, int m) {
  int d[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  return d[m] + (y % 4 == 0);
}
int main() {
  cin >> in;
  map<string, int> cnt;
  for (int i = 0; i < (int)in.size() - 9; i++) {
    if (isdigit(in[i]) && isdigit(in[i + 1]) && isdigit(in[i + 3]) &&
        isdigit(in[i + 4]) && isdigit(in[i + 6]) && isdigit(in[i + 7]) &&
        isdigit(in[i + 8]) && isdigit(in[i + 9]) && in[i + 2] == '-' &&
        in[i + 5] == '-') {
      int y = atoi(in.substr(i + 6, 4).c_str());
      int m = atoi(in.substr(i + 3, 2).c_str());
      int d = atoi(in.substr(i, 2).c_str());
      if (2013 <= y && y <= 2015 && 1 <= m && m <= 12 && 1 <= d &&
          d <= day(y, m))
        cnt[in.substr(i, 10)]++;
    }
  }
  int mx = 0;
  for (__typeof(cnt.begin()) i = cnt.begin(); i != cnt.end(); i++)
    mx = max(mx, i->second);
  for (__typeof(cnt.begin()) i = cnt.begin(); i != cnt.end(); i++)
    if (i->second == mx) cout << i->first << endl;
  return 0;
}
