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;
int d, m, y, k, mx, mxi;
int day[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool f;
int dd[5000], mm[5000], yy[5000], s[5000];
int pm(char x) {
  if (((x - 48) >= 0) && ((x - 48) <= 9)) {
    return x - 48;
  } else {
    return -1;
  }
}
int main() {
  string a;
  cin >> a;
  for (int i = 0; i < a.length(); i++) {
    if ((pm(a[i]) != -1) && (pm(a[i + 1]) != -1) && (a[i + 2] == '-') &&
        (pm(a[i + 3]) != -1) && (pm(a[i + 4]) != -1) && (a[i + 5] == '-') &&
        (pm(a[i + 6]) != -1) && (pm(a[i + 7]) != -1) && (pm(a[i + 8]) != -1) &&
        (pm(a[i + 9]) != -1) && (pm(a[i + 3]) * 10 + pm(a[i + 4]) <= 12) &&
        ((pm(a[i]) * 10 + pm(a[i + 1]) <=
          day[pm(a[i + 3]) * 10 + pm(a[i + 4])]))) {
      d = pm(a[i]) * 10 + pm(a[i + 1]);
      m = pm(a[i + 3]) * 10 + pm(a[i + 4]);
      y = pm(a[i + 6]) * 1000 + pm(a[i + 7]) * 100 + pm(a[i + 8]) * 10 +
          pm(a[i + 9]);
      if ((d == 0) || (m == 0)) continue;
      if ((y < 2013) || (y > 2015)) continue;
      f = true;
      for (int j = 0; j < k; j++) {
        if ((dd[j] == d) && (mm[j] == m) && (yy[j] == y)) {
          f = false;
          s[j]++;
        }
      }
      if (f) {
        dd[k] = d;
        mm[k] = m;
        yy[k] = y;
        s[k] = 1;
        k++;
      }
    }
  }
  for (int i = 0; i < k; i++) {
    if (s[i] > mx) {
      mx = s[i];
      mxi = i;
    }
  }
  if (dd[mxi] / 10 == 0) {
    cout << 0 << dd[mxi] << "-";
  } else {
    cout << dd[mxi] << "-";
  }
  if (mm[mxi] / 10 == 0) {
    cout << 0 << mm[mxi] << "-";
  } else {
    cout << mm[mxi] << "-";
  }
  cout << yy[mxi];
  return 0;
}
