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;
bool prime[100000 + 1];
void sieve() {
  memset(prime, true, sizeof(prime));
  for (long p = 2; p * p <= 100000; p++) {
    if (prime[p] == true) {
      for (long i = p * p; i <= 100000; i += p) prime[i] = false;
    }
  }
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long power(int v, int a) {
  long prod = 1;
  for (int i = 0; i < a; i++) prod *= v;
  return prod;
}
int countFreq(string &pat, string &txt) {
  int M = pat.length();
  int N = txt.length();
  int res = 0;
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (txt[i + j] != pat[j]) break;
    if (j == M) {
      res++;
      j = 0;
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T = 1;
  while (T--) {
    string s;
    vector<string> v;
    int d[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    for (int i = 2013; i <= 2015; i++) {
      for (int j = 1; j <= 12; j++) {
        for (int k = 1; k <= d[j - 1]; k++) {
          string temp = "";
          if (k / 10 == 0) temp += "0";
          temp += to_string(k) + "-";
          if (j / 10 == 0) temp += "0";
          temp += to_string(j) + "-";
          temp += to_string(i);
          v.push_back(temp);
        }
      }
    }
    cin >> s;
    int ans = 0;
    string final;
    for (int i = 0; i < v.size(); i++) {
      int t = countFreq(v[i], s);
      if (ans < t) {
        ans = t;
        final = v[i];
      }
    }
    cout << final;
  }
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}
