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;
ifstream fin("input.txt");
ofstream fout("output.txt");
int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int cnt[100000];
int calc(string str) {
  if (str[2] != '-' || str[5] != '-') return 50000;
  if (str[6] != '2') return 50000;
  if (str[7] != '0') return 50000;
  if (str[8] != '1') return 50000;
  if (str[9] < '3' || str[9] > '5') return -1;
  for (int i = 0; i < 5; i++)
    if (str[i] == '-' && i != 2) return 50000;
  int d = (str[0] - '0') * 10 + str[1] - '0';
  int m = (str[3] - '0') * 10 + str[4] - '0';
  if (m < 1 || m > 12) return 50000;
  if (d < 1 || d > day[m - 1]) return 50000;
  return (str[9] - '3') * 10000 + m * 100 + d;
}
int main() {
  string str;
  cin >> str;
  for (int i = 0; i + 9 < str.length(); i++) cnt[calc(str.substr(i, 10))]++;
  int idx = -1, mx = 0;
  for (int i = 0; i < 30000; i++)
    if (mx < cnt[i]) idx = i, mx = cnt[i];
  printf("%02d-%02d-%d\n", idx % 100, (idx / 100) % 100, 2013 + idx / 10000);
  return 0;
}
