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 n;
long long table[40][40];
long long solve(int nodes, int h) {
  if (nodes == 0) {
    return 1LL;
  }
  if (h == 0) {
    return 0;
  }
  long long &ret = table[nodes][h];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < nodes; ++i) {
    ret += solve(i, max(h - 1, 0)) * solve(nodes - i - 1, max(h - 1, 0));
  }
  return ret;
}
string days[] = {"01", "02", "03", "04", "05", "06", "07", "08",
                 "09", "10", "11", "12", "13", "14", "15", "16",
                 "17", "18", "19", "20", "21", "22", "23", "24",
                 "25", "26", "27", "28", "29", "30", "31"};
string months[] = {"01", "02", "03", "04", "05", "06",
                   "07", "08", "09", "10", "11", "12"};
int lim[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int fail[100];
char t[100010], p[100];
inline void buildFail() {
  int m = strlen(p);
  int j = fail[0] = -1;
  for (int i = 1; i <= m; ++i) {
    while (j >= 0 && p[j] != p[i - 1]) j = fail[j];
    fail[i] = ++j;
  }
}
int match() {
  int n = strlen(t), m = strlen(p);
  int count = 0;
  for (int i = 0, k = 0; i < n; ++i) {
    while (k >= 0 && p[k] != t[i]) k = fail[k];
    if (++k >= m) {
      ++count;
      k = fail[k];
    }
  }
  return count;
}
int main() {
  string in;
  cin >> t;
  int maxo = 0, koko;
  string ans;
  string tmpo;
  for (int i = 0; i < 12; ++i) {
    for (int j = 0; j < lim[i]; ++j) {
      tmpo = days[j] + "-" + months[i] + "-" + "2013";
      strcpy(p, tmpo.c_str());
      buildFail();
      koko = match();
      if (koko > maxo) {
        maxo = koko;
        ans = tmpo;
      }
      tmpo = days[j] + "-" + months[i] + "-" + "2014";
      strcpy(p, tmpo.c_str());
      buildFail();
      koko = match();
      if (koko > maxo) {
        maxo = koko;
        ans = tmpo;
      }
      tmpo = days[j] + "-" + months[i] + "-" + "2015";
      strcpy(p, tmpo.c_str());
      buildFail();
      koko = match();
      if (koko > maxo) {
        maxo = koko;
        ans = tmpo;
      }
    }
  }
  cout << ans << endl;
  return 0;
}
