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;
map<string, int> sum;
string ans;
int maks = -1;
int pola[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
char in[100008];
int len;
bool valid(int a, int b) {
  for (int i = a; i <= b; i++) {
    if (i == a + 2 || i == a + 5) {
      if (in[i] != '-') return 0;
    } else {
      if (in[i] == '-') return 0;
    }
  }
  int day = ((in[a] - '0') * 10) + (in[a + 1] - '0');
  int month = ((in[a + 3] - '0') * 10) + (in[a + 4] - '0');
  int year = ((in[a + 6] - '0') * 1000) + ((in[a + 7] - '0') * 100) +
             ((in[a + 8] - '0') * 10) + (in[a + 9] - '0');
  if (year < 2013 || year > 2015) return 0;
  if (month < 1 || month > 13 || day < 1) return 0;
  if (day > pola[month - 1]) return 0;
  return 1;
}
void process() {
  int a = 0, b = 9;
  while (b < len) {
    if (valid(a, b)) {
      string tmp = "";
      for (int i = a; i <= b; i++) {
        tmp += in[i];
      }
      sum[tmp]++;
      int n = sum[tmp];
      if (n > maks) {
        maks = n;
        ans = tmp;
      }
    }
    a++;
    b++;
  }
hell:
  cout << ans << endl;
}
int main() {
  scanf("%s", in);
  len = strlen(in);
  process();
}
