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 maxn = 100010;
const int d[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int t, n;
string s;
map<int, int> red;
void output(int idx, int a[]) {
  for (int i = 0; i <= idx; i++) cout << a[i] << endl;
}
void work() {
  red.clear();
  int idx = 0;
  string res[maxn], a = "";
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '-') {
      res[idx++] = a;
      a = "";
    } else {
      a.push_back(s[i]);
    }
  }
  res[idx] = a;
  for (int i = idx; i >= 2; i--) {
    if (res[i].length() >= 4) {
      int tmp = (res[i][0] - '0') * 1000 + (res[i][1] - '0') * 100 +
                (res[i][2] - '0') * 10 + (res[i][3] - '0');
      if (tmp >= 2013 && tmp <= 2015) {
        int dd = tmp * 10000;
        if (res[i - 1].length() == 2) {
          tmp = (res[i - 1][0] - '0') * 10 + (res[i - 1][1] - '0');
          if (tmp > 0 && tmp <= 12) {
            dd += tmp * 100;
            if (res[i - 2].length() >= 2) {
              int mm = tmp;
              tmp = (res[i - 2][res[i - 2].length() - 2] - '0') * 10 +
                    (res[i - 2][res[i - 2].length() - 1] - '0');
              if (tmp > 0 && tmp <= d[mm]) {
                dd += tmp;
                red[dd]++;
              }
            }
          }
        }
      }
    }
  }
  int amt = 0, ans;
  for (map<int, int>::iterator it = red.begin(); it != red.end(); it++) {
    if (it->second >= amt) {
      amt = it->second;
      ans = it->first;
    }
  }
  for (int i = 0; i < 2; i++) {
    if (ans % 100 < 10)
      cout << "0" << ans % 100 << "-";
    else
      cout << ans % 100 << "-";
    ans /= 100;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  while (cin >> s) work();
  return 0;
}
