#include <bits/stdc++.h>
using namespace std;
bool tep(pair<string, int> a, pair<string, int> b) {
  return a.second > b.second;
}
bool f(string s) {
  for (int i = 0; i < 10; i++) {
    if ((i == 2 || i == 5) && s[i] != '-') return 0;
    if ((i != 2 && i != 5) && s[i] == '-') return 0;
  }
  if (s.substr(6) != "2013" && s.substr(6) != "2014" && s.substr(6) != "2015")
    return 0;
  if ((s[3] != '0' && s[4] != '0' && s[4] != '1' && s[4] != '2') ||
      s[3] > '1' || (s[3] == '0' && s[4] == '0'))
    return 0;
  if (s.substr(0, 2) == "00") return 0;
  if (s.substr(3, 2) == "02" && (s[0] >= '3' || (s[0] == '2' && s[1] > '8')))
    return 0;
  if ((s.substr(3, 2) == "04" || s.substr(3, 2) == "06" ||
       s.substr(3, 2) == "09" || s.substr(3, 2) == "11") &&
      (s[0] > '3' || (s[0] == '3' && s[1] > '0')))
    return 0;
  else if (s.substr(3, 2) != "02" &&
           (s[0] > '3' || (s[0] == '3' && s[1] > '1')))
    return 0;
  return 1;
}
int main() {
  string str;
  vector<pair<string, int> > x;
  int xn = 0;
  cin >> str;
  for (int i = 0; i < str.length() - 9; i++) {
    string ss = str.substr(i, 10);
    if (f(ss)) {
      int j;
      for (j = 0; j < xn; j++)
        if (ss == x[j].first) {
          x[j].second++;
          break;
        }
      if (j == xn) {
        xn++;
        x.push_back(make_pair(ss, 1));
      }
    }
  }
  pair<string, int> xx[xn];
  for (int i = 0; i < xn; i++) {
    xx[i].first = x[i].first;
    xx[i].second = x[i].second;
  }
  sort(xx, xx + xn, tep);
  cout << xx[0].first << endl;
  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;
bool tep(pair<string, int> a, pair<string, int> b) {
  return a.second > b.second;
}
bool f(string s) {
  for (int i = 0; i < 10; i++) {
    if ((i == 2 || i == 5) && s[i] != '-') return 0;
    if ((i != 2 && i != 5) && s[i] == '-') return 0;
  }
  if (s.substr(6) != "2013" && s.substr(6) != "2014" && s.substr(6) != "2015")
    return 0;
  if ((s[3] != '0' && s[4] != '0' && s[4] != '1' && s[4] != '2') ||
      s[3] > '1' || (s[3] == '0' && s[4] == '0'))
    return 0;
  if (s.substr(0, 2) == "00") return 0;
  if (s.substr(3, 2) == "02" && (s[0] >= '3' || (s[0] == '2' && s[1] > '8')))
    return 0;
  if ((s.substr(3, 2) == "04" || s.substr(3, 2) == "06" ||
       s.substr(3, 2) == "09" || s.substr(3, 2) == "11") &&
      (s[0] > '3' || (s[0] == '3' && s[1] > '0')))
    return 0;
  else if (s.substr(3, 2) != "02" &&
           (s[0] > '3' || (s[0] == '3' && s[1] > '1')))
    return 0;
  return 1;
}
int main() {
  string str;
  vector<pair<string, int> > x;
  int xn = 0;
  cin >> str;
  for (int i = 0; i < str.length() - 9; i++) {
    string ss = str.substr(i, 10);
    if (f(ss)) {
      int j;
      for (j = 0; j < xn; j++)
        if (ss == x[j].first) {
          x[j].second++;
          break;
        }
      if (j == xn) {
        xn++;
        x.push_back(make_pair(ss, 1));
      }
    }
  }
  pair<string, int> xx[xn];
  for (int i = 0; i < xn; i++) {
    xx[i].first = x[i].first;
    xx[i].second = x[i].second;
  }
  sort(xx, xx + xn, tep);
  cout << xx[0].first << endl;
  return 0;
}

```