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[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int o[31][12][3];
bool isDigit(char c) { return '0' <= c && c <= '9'; }
int toInt(char c) { return c - '0'; }
int main() {
  char t[100001];
  scanf("%s", t);
  int l = strlen(t);
  for (int i = 0; i <= l - 10; i++) {
    if (isDigit(t[i]) && isDigit(t[i + 1]) && isDigit(t[i + 3]) &&
        isDigit(t[i + 4]) && isDigit(t[i + 6]) && isDigit(t[i + 7]) &&
        isDigit(t[i + 8]) && isDigit(t[i + 9]))
      if (t[i + 2] == '-' && t[i + 5] == '-') {
        int a = 10 * toInt(t[i]) + toInt(t[i + 1]);
        int b = 10 * toInt(t[i + 3]) + toInt(t[i + 4]);
        int c = 1000 * toInt(t[i + 6]) + 100 * toInt(t[i + 7]) +
                10 * toInt(t[i + 8]) + toInt(t[i + 9]);
        if (1 <= b && b <= 12)
          if (1 <= a && a <= d[b - 1])
            if (2013 <= c && c <= 2015) o[a - 1][b - 1][c - 2013]++;
      }
  }
  int i = 0, j = 0, k = 0;
  for (int a = 0; a < 31; a++)
    for (int b = 0; b < 12; b++)
      for (int c = 0; c < 3; c++)
        if (o[a][b][c] > o[i][j][k]) {
          i = a;
          j = b;
          k = c;
        }
  printf("%s%d-%s%d-%d\n", i + 1 < 10 ? "0" : "", i + 1, j + 1 < 10 ? "0" : "",
         j + 1, k + 2013);
  return 0;
}
