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;
string s;
int num[2][13][105];
int month[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
void solve(string d) {
  if (d[0] == '-' || d[1] == '-' || d[2] != '-' || d[3] == '-' || d[4] == '-' ||
      d[5] != '-' || d[6] == '-' || d[7] == '-' || d[8] == '-' || d[9] == '-')
    return;
  int p = 0, y = 0, m = 0;
  string t;
  for (int i = 0; i < 2; i++) p = p * 10 + d[i] - '0';
  for (int i = 3; i < 5; i++) m = m * 10 + d[i] - '0';
  for (int i = 6; i < 10; i++) y = y * 10 + d[i] - '0';
  if (y < 2013 || y > 2015) return;
  if (m < 1 || m > 12) return;
  if (p < 1 || p > month[m]) return;
  num[y - 2013][m][p]++;
}
int y, m, d, mx;
void print() {
  if (d < 10) cout << '0';
  cout << d << '-';
  if (m < 10) cout << '0';
  cout << m;
  cout << '-' << y;
}
int main() {
  cin >> s;
  for (int i = 0; i + 9 < s.size(); i++) solve(s.substr(i, 10));
  for (int i = 0; i < 3; i++)
    for (int j = 1; j <= 12; j++)
      for (int k = 1; k <= month[j]; k++) {
        if (num[i][j][k] > mx) {
          y = i + 2013, m = j, d = k;
          mx = num[i][j][k];
        }
      }
  print();
  return 0;
}
