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>
namespace patch {
template <typename T>
std::string to_string(const T& n) {
  std::ostringstream stm;
  stm << n;
  return stm.str();
}
}  // namespace patch
using namespace std;
long long int days[13] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
map<string, long long int> mp;
bool chk(char a) { return (a >= '0' && a <= '9'); }
void solve(string S) {
  if (S[2] == '-' && S[5] == S[2] && chk(S[0]) && chk(S[1]) && chk(S[3]) &&
      chk(S[4]) && chk(S[6]) && chk(S[7]) && chk(S[8]) && chk(S[9])) {
    stringstream ss;
    ss << S;
    long long int D = 0, M = 0, Y = 0;
    ss >> D >> M >> Y;
    D = abs(D);
    M = abs(M);
    Y = abs(Y);
    if (D == 0 || D > 31 || M == 0 || M > 12 || Y < 2013 || Y > 2015) {
      return;
    }
    if (D > 0 && D <= days[M - 1]) {
      mp[S]++;
    }
  }
}
int main() {
  string S;
  cin >> S;
  for (int i = 0; i < S.size() - 9; i++) {
    string tmp = "";
    for (int j = i; j < i + 10; j++) {
      tmp += S[j];
    }
    solve(tmp);
  }
  long long int mx = 0;
  string tmp;
  for (auto it : mp) {
    if (mx < it.second) {
      mx = it.second;
      tmp = it.first;
    }
  }
  stringstream ss;
  long long int X;
  ss << tmp;
  ss >> X;
  X = abs(X);
  if (X < 10) {
    cout << 0;
  }
  cout << X << '-';
  ss >> X;
  X = abs(X);
  if (X < 10) {
    cout << 0;
  }
  cout << X << '-';
  ss >> X;
  cout << abs(X);
  return 0;
}
