#include <bits/stdc++.h>
using namespace std;
double pi = 3.141592653589793238;
const int M = 1e9 + 7;
const int Nmax = 5005;
const int MM = 1e7 + 1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    set<string> ss;
    string s;
    cin >> s;
    int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    for (int i = 2013; i <= 2015; i++) {
      for (int j = 1; j <= 12; j++) {
        for (int k = 1; k <= day[j - 1]; k++) {
          string date = "";
          if (k < 10) date += '0';
          date += to_string(k);
          date += '-';
          if (j < 10) date += '0';
          date += to_string(j);
          date += '-';
          date += to_string(i);
          ss.insert(date);
        }
      }
    }
    map<string, int> m;
    for (int i = 0; i <= s.size() - 10; i++) {
      string k = s.substr(i, 10);
      if (ss.count(k)) m[k]++;
    }
    int mx = 0;
    string ans;
    for (auto x : m) {
      if (x.second > mx) {
        ans = x.first;
        mx = x.second;
      }
    }
    cout << ans;
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
double pi = 3.141592653589793238;
const int M = 1e9 + 7;
const int Nmax = 5005;
const int MM = 1e7 + 1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    set<string> ss;
    string s;
    cin >> s;
    int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    for (int i = 2013; i <= 2015; i++) {
      for (int j = 1; j <= 12; j++) {
        for (int k = 1; k <= day[j - 1]; k++) {
          string date = "";
          if (k < 10) date += '0';
          date += to_string(k);
          date += '-';
          if (j < 10) date += '0';
          date += to_string(j);
          date += '-';
          date += to_string(i);
          ss.insert(date);
        }
      }
    }
    map<string, int> m;
    for (int i = 0; i <= s.size() - 10; i++) {
      string k = s.substr(i, 10);
      if (ss.count(k)) m[k]++;
    }
    int mx = 0;
    string ans;
    for (auto x : m) {
      if (x.second > mx) {
        ans = x.first;
        mx = x.second;
      }
    }
    cout << ans;
  }
  return 0;
}

```