#include <bits/stdc++.h>
using namespace std;
long long int biexp(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b % 2) res = res * a;
    a = a * a;
    b = b / 2;
  }
  return res;
}
int main() {
  int t = 1;
  while (t--) {
    string s;
    cin >> s;
    string r;
    long long int x = 0;
    string s2;
    map<string, long long int> v;
    for (long long int i = 0; i <= s.length() - 10; i++) {
      r = s.substr(i, 10);
      long long int d = (r[0] - '0') * 10 + (r[1] - '0');
      long long int m = (r[3] - '0') * 10 + (r[4] - '0');
      if (r[0] != '-' and r[1] != '-' and r[2] == '-' and r[3] != '-' and
          r[4] != '-' and r[5] == '-' and r[6] == '2' and r[7] == '0' and
          r[8] == '1' and (r[9] == '3' or r[9] == '4' or r[9] == '5')) {
        if (m >= 1 and m <= 12) {
          if (m == 2) {
            if (d >= 1 and d <= 28) v[r]++;
          } else if (((m <= 7 and m % 2 != 0) or (m >= 8 and m % 2 == 0)) and
                     d >= 1 and d <= 31)
            v[r]++;
          else if (d >= 1 and d <= 30)
            v[r]++;
        }
      }
    }
    for (auto it = v.begin(); it != v.end(); it++) {
      if (it->second > x) {
        x = it->second;
        s2 = it->first;
      }
    }
    cout << s2;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int biexp(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b % 2) res = res * a;
    a = a * a;
    b = b / 2;
  }
  return res;
}
int main() {
  int t = 1;
  while (t--) {
    string s;
    cin >> s;
    string r;
    long long int x = 0;
    string s2;
    map<string, long long int> v;
    for (long long int i = 0; i <= s.length() - 10; i++) {
      r = s.substr(i, 10);
      long long int d = (r[0] - '0') * 10 + (r[1] - '0');
      long long int m = (r[3] - '0') * 10 + (r[4] - '0');
      if (r[0] != '-' and r[1] != '-' and r[2] == '-' and r[3] != '-' and
          r[4] != '-' and r[5] == '-' and r[6] == '2' and r[7] == '0' and
          r[8] == '1' and (r[9] == '3' or r[9] == '4' or r[9] == '5')) {
        if (m >= 1 and m <= 12) {
          if (m == 2) {
            if (d >= 1 and d <= 28) v[r]++;
          } else if (((m <= 7 and m % 2 != 0) or (m >= 8 and m % 2 == 0)) and
                     d >= 1 and d <= 31)
            v[r]++;
          else if (d >= 1 and d <= 30)
            v[r]++;
        }
      }
    }
    for (auto it = v.begin(); it != v.end(); it++) {
      if (it->second > x) {
        x = it->second;
        s2 = it->first;
      }
    }
    cout << s2;
  }
  return 0;
}

```