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 a;
int h1[100][100][10];
int moth[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool panduan[10] = {0, 0, 1, 0, 0, 1};
bool pan(string m) {
  string h(m, 6, 4);
  if (h != "2013" && h != "2014" && h != "2015") return 0;
  int u = (m[0] - '0') * 10 + m[1] - '0', v = (m[3] - '0') * 10 + m[4] - '0',
      z = m[9] - '0';
  if ((0 < v && v <= 12 && u > 0 && u <= moth[v]) == 0) return 0;
  for (int i = 0; i < 10; i++)
    if (panduan[i] != (m[i] == '-')) return 0;
  h1[u][v][z]++;
  return 1;
}
string huanyuan(int i, int j, int o) {
  string u = "";
  u += char(i / 10 + int('0'));
  u += char(i % 10 + int('0'));
  u += '-';
  u += char(j / 10 + int('0'));
  u += char(j % 10 + int('0'));
  u += '-';
  u += "201";
  u += char(o + int('0'));
  return u;
}
int main(int argc, char *argv[]) {
  cin >> a;
  for (int i = 0; i <= a.size() - 10; i++) {
    string m(a, i, 10);
    pan(m);
  }
  string e;
  int u = 0;
  for (int i = 0; i < 40; i++) {
    for (int j = 0; j < 20; j++) {
      for (int o = 0; o < 10; o++) {
        if (h1[i][j][o] > u) {
          u = h1[i][j][o];
          e = huanyuan(i, j, o);
        }
      }
    }
  }
  cout << e << endl;
  return 0;
}
