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;
const int maxn = 1e5 + 10;
int arr[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
char str[maxn];
map<string, int> mp;
struct CmpByValue {
  bool operator()(const pair<string, int>& lhs, const pair<string, int>& rhs) {
    return lhs.second > rhs.second;
  }
};
int check(int x) {
  int flag[5] = {0}, numd = 0, numm = 0, numy = 0;
  if (isdigit(str[x - 1]) && isdigit(str[x - 2]))
    flag[0] = 1, numd = (str[x - 2] - '0') * 10 + (str[x - 1] - '0');
  if (isdigit(str[x + 1]) && isdigit(str[x + 2]))
    flag[1] = 1, numm = (str[x + 1] - '0') * 10 + (str[x + 2] - '0');
  for (int i = x + 4; i < x + 8; i++)
    if (isdigit(str[i]))
      numy = numy * 10 + (str[i] - '0');
    else
      return 0;
  if (flag[0] && flag[1]) {
    if (2013 <= numy && numy <= 2015) {
      if (1 <= numm && numm <= 12) {
        if (0 < numd && numd <= arr[numm])
          return 1;
        else
          return 0;
      } else
        return 0;
    } else
      return 0;
  } else
    return 0;
}
int main() {
  scanf("%s", str);
  int n = strlen(str);
  for (int i = 0; i < n; i++) {
    if (str[i] == '-' && str[i + 3] == '-') {
      if (i - 2 < 0 || i + 7 >= n) continue;
      if (check(i)) {
        string s;
        for (int j = i - 2; j <= i + 7; j++) s += str[j];
        mp[s]++;
      }
    }
  }
  vector<pair<string, int> > vec(mp.begin(), mp.end());
  sort(vec.begin(), vec.end(), CmpByValue());
  pair<string, int> p = vec[0];
  cout << p.first << endl;
  return 0;
}
