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 d[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
string ans[1000010];
int a[40000000], b[1000000];
int main() {
  string str;
  int ma = 0, tot = 0, t = 0, an = 0;
  cin.tie(0);
  cin >> str;
  for (int i = 2; i < str.length() - 6; ++i)
    if (str[i] == '-' && str[i + 3] == '-' && str[i - 1] != '-' &&
        str[i + 1] != '-' && str[i + 7] != '-' && str[i + 2] != '-' &&
        str[i - 2] != '-') {
      tot++;
      for (int j = i - 2; j <= i + 7; ++j) ans[tot] += str[j];
    }
  for (int i = 1; i <= tot; ++i) {
    int sum1, sum2, sum3;
    sum1 = (ans[i][0] - '0') * 10 + (ans[i][1] - '0');
    sum2 = (ans[i][3] - '0') * 10 + (ans[i][4] - '0');
    sum3 = (ans[i][6] - '0') * 1000 + (ans[i][7] - '0') * 100 +
           (ans[i][8] - '0') * 10 + (ans[i][9] - '0');
    if (sum1 > d[sum2]) continue;
    if (sum3 < 2013 || sum3 > 2015) continue;
    if (sum1 <= 0) continue;
    a[sum3 * 10000 + sum2 * 100 + sum1]++;
    b[++t] = sum3 * 10000 + sum2 * 100 + sum1;
  }
  for (int i = 1; i <= t; ++i) {
    if (a[b[i]] > ma) ma = a[b[i]], an = b[i];
  }
  if (an % 100 < 10)
    cout << 0 << an % 100 << '-';
  else
    cout << an % 100 << '-';
  if (an / 100 % 100 < 10)
    cout << 0 << an / 100 % 100 << '-';
  else
    cout << an / 100 % 100 << '-';
  cout << an / 10000;
}
