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 inf = 1000000000;
const long long linf = 1000000000000000000LL;
typedef struct {
  int dia, mes, ano;
} data;
vector<data> vetor;
string linha, str;
map<int, int> mapa;
int main() {
  cin >> linha;
  int s = linha.size();
  data d;
  for (int i = int(0); i <= int(s - 10); i++) {
    str = linha.substr(i, 10);
    if (isdigit(str[0]) && isdigit(str[1]) && !isdigit(str[2]) &&
        isdigit(str[3]) && isdigit(str[4]) && !isdigit(str[5]) &&
        isdigit(str[6]) && isdigit(str[7]) && isdigit(str[8]) &&
        isdigit(str[9])) {
      d.dia = (str[0] - '0') * 10 + str[1] - '0';
      d.mes = (str[3] - '0') * 10 + str[4] - '0';
      d.ano = (str[6] - '0') * 1000 + (str[7] - '0') * 100 +
              (str[8] - '0') * 10 + str[9] - '0';
      if (d.ano >= 2013 && d.ano <= 2015 && d.mes >= 1 && d.mes <= 12 &&
          d.dia) {
        if (d.mes == 2) {
          if (d.dia <= 28) vetor.push_back(d);
        } else if (d.mes <= 7) {
          if (d.mes % 2) {
            if (d.dia <= 31) vetor.push_back(d);
          } else if (d.dia <= 30)
            vetor.push_back(d);
        } else {
          if (d.mes % 2) {
            if (d.dia <= 30) vetor.push_back(d);
          } else if (d.dia <= 31)
            vetor.push_back(d);
        }
      }
      mapa[1000000 * d.dia + 10000 * d.mes + d.ano] = 0;
    }
  }
  s = vetor.size();
  int a, b, c;
  for (int i = 0; i < s; i++) {
    a = vetor[i].dia, b = vetor[i].mes, c = vetor[i].ano;
    mapa[1000000 * a + 10000 * b + c]++;
  }
  int m = 0;
  for (map<int, int>::iterator it = mapa.begin(); it != mapa.end(); it++)
    if ((it->second) > m) m = it->second, a = it->first;
  printf("%.2d-%.2d-%.4d\n", a / 1000000, a % 1000000 / 10000, a % 10000);
  return 0;
}
