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;
struct Node {
  int y, m, d;
  Node(int y = 0, int m = 0, int d = 0) : y(y), m(m), d(d) {}
  bool operator<(const Node &b) const {
    if (y != b.y) return y < b.y;
    if (m != b.m) return m < b.m;
    return d < b.d;
  }
  void out() { printf("%02d-%02d-%04d\n", d, m, y); }
};
const int b[8] = {0, 1, 3, 4, 6, 7, 8, 9};
const int f[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
char s[200005];
map<Node, int> mp;
inline int val(char c) { return c - '0'; }
inline int day(char *a) { return val(a[0]) * 10 + val(a[1]); }
inline int month(char *a) { return val(a[0]) * 10 + val(a[1]); }
inline int year(char *a) { return day(a) * 100 + day(a + 2); }
void gao(char *a) {
  int y, m, d;
  for (int i = 0; i < 8; ++i)
    if (!isdigit(a[b[i]])) return;
  if (a[2] != '-' || a[5] != '-') return;
  d = day(a);
  m = month(a + 3);
  if (m < 1 || m > 12) return;
  if (d < 1 || d > f[m]) return;
  y = year(a + 6);
  if (y < 2013 || y > 2015) return;
  ++mp[Node(y, m, d)];
}
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i + 10 <= n; ++i) gao(s + i);
  Node ans;
  int cnt = -1;
  for (map<Node, int>::iterator it = mp.begin(); it != mp.end(); ++it)
    if (it->second > cnt) {
      cnt = it->second;
      ans = it->first;
    }
  ans.out();
  return 0;
}
