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 = 1e9 + 7;
string s;
map<string, int> cnt;
int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
string ans = "0";
void solve(string x) {
  if (x[2] != '-' or x[5] != '-') return;
  for (int i = 0; i < 10; i++) {
    if ((i == 2 or i == 5) and x[i] != '-') return;
    if (i == 2 or i == 5) continue;
    if (x[i] < '0' or x[i] > '9') return;
  }
  int day = (x[0] - '0') * 10 + x[1] - '0';
  int month = (x[3] - '0') * 10 + x[4] - '0';
  int year =
      (x[6] - '0') * 1000 + (x[7] - '0') * 100 + (x[8] - '0') * 10 + x[9] - '0';
  if (year < 2013 or year > 2015) return;
  if (month < 1 or month > 12) return;
  if (day < 1 or day > days[month]) return;
  if (cnt.find(x) == cnt.end())
    cnt[x] = 1;
  else
    cnt[x]++;
  if (cnt[x] > cnt[ans]) ans = x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  cnt["0"] = -1;
  for (int i = 0; i + 9 < s.size(); i++) {
    solve(s.substr(i, 10));
  }
  cout << ans;
  return 0;
}
