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 a[2000] = {};
string r[2000];
const int mon[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
void tr(int p) {
  int res = true;
  for (int i = 0; res && i < 10; i++)
    res = (((i == 2 || i == 5) && s[p + i] == '-') ||
           (i != 2 && i != 5 && s[p + i] != '-'));
  if (!res) return;
  int d = (s[p] - '0') * 10 + s[p + 1] - '0',
      m = (s[p + 3] - '0') * 10 + s[p + 4] - '0',
      y = (((s[p + 6] - '0') * 10 + s[p + 7] - '0') * 10 + s[p + 8] - '0') *
              10 +
          s[p + 9] - '0';
  if (y >= 2013 && y <= 2015 && m >= 1 && m <= 12 && d >= 1 &&
      d <= mon[m - 1]) {
    d += ((y - 2013) * 12 + m - 1) * 31;
    a[d]++;
    r[d] = "";
    for (int i = 0; i < 10; i++) r[d] += s[i + p];
  }
}
int main() {
  cin >> s;
  for (int i = 0; i <= s.size() - 10; i++) tr(i);
  int ans = 0;
  for (int i = 0; i < 1500; i++)
    if (a[i] > a[ans]) ans = i;
  cout << r[ans];
}
