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;
int month[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int ans[33][33][2222], n, k;
string t, res;
int dig(char c) { return (int)(c - '0'); }
void check() {
  if (t[2] != '-' || t[5] != '-') return;
  int cnt = 0;
  for (int i = 0; i < 10; i++)
    if (t[i] == '-') cnt++;
  if (cnt > 2) return;
  int d = dig(t[0]) * 10 + dig(t[1]);
  int m = dig(t[3]) * 10 + dig(t[4]);
  int y = dig(t[6]) * 1000 + dig(t[7]) * 100 + dig(t[8]) * 10 + dig(t[9]);
  if (m < 1 || m > 12) return;
  if (d < 1 || d > month[m]) return;
  if (y < 2013 || y > 2015) return;
  ans[d][m][y]++;
  if (ans[d][m][y] > k) {
    k = ans[d][m][y];
    res = t;
  }
}
int main() {
  string s;
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) {
    t = "00-00-0000";
    for (int j = i, at = 0; j < i + 10 && j < n; j++, at++) t[at] = s[j];
    check();
  }
  cout << res << endl;
  return 0;
}
