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 days[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int ok[] = {0, 1, 3, 4, 6, 7, 8, 9};
int len;
map<string, int> m;
char in[100001];
void solve(int p) {
  if (p + 10 > len) return;
  string aux;
  for (int i = 0; i < 10; ++i) aux += in[i + p];
  bool k = 1;
  for (int i = 0; i < 8 && k; ++i) k = aux[ok[i]] <= '9' && aux[ok[i]] >= '0';
  k &= aux[2] == '-' && aux[5] == '-';
  if (k) {
    int day, month, year;
    sscanf(aux.c_str(), "%d-%d-%d", &day, &month, &year);
    if (month > 0 && month <= 12 && day > 0 && day <= days[month] &&
        year >= 2013 && year <= 2015)
      m[aux]++;
  }
  solve(p + 1);
}
int main() {
  scanf("%s", in);
  len = strlen(in);
  solve(0);
  vector<pair<int, string> > vet;
  for (typeof(m.begin()) it = m.begin(); it != m.end(); ++it)
    vet.push_back(make_pair(it->second, it->first));
  sort(vet.begin(), vet.end());
  printf("%s\n", vet[vet.size() - 1].second.c_str());
}
