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;
map<string, int> mp;
const int day[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n = 0, y, m, d, j;
string str, v, x;
int main() {
  cin >> str;
  for (int i = 0; i + 10 <= str.length(); i++) {
    x = str.substr(i, 10);
    if (sscanf((x + "#1").c_str(), "%2d-%2d-%4d#%d", &d, &m, &y, &j) != 4)
      continue;
    if (j != 1 || y < 2013 || y > 2015 || m < 1 || m > 12 || d < 1 ||
        d > day[m])
      continue;
    mp[x]++;
    if (n < mp[x]) n = mp[x], v = x;
  }
  cout << v << endl;
  return 0;
}
