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 day[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int record[5][20][40];
int main() {
  char s[100015];
  vector<int> num[3];
  int len, m, d, y, ans, M, D, Y;
  while (scanf(" %s", s) != EOF) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 20; j++)
        for (int k = 0; k < 40; k++) record[i][j][k] = 0;
    len = strlen(s), ans = 0;
    for (int i = 0; i < 3; i++) num[i].clear();
    s[len++] = '-';
    for (int i = 0; i < len; i++) {
      if (s[i] == '-') {
        if (num[0].size() > 0 && num[1].size() == 2 && num[2].size() > 0) {
          if (num[0].size() >= 2)
            d = num[0][num[0].size() - 2] * 10 + num[0][num[0].size() - 1];
          m = num[1][0] * 10 + num[1][1];
          y = 0;
          for (int x = 0; x < 4; x++) y = y * 10 + num[2][x];
          if (2013 <= y && y <= 2015 && 1 <= m && m <= 12 && 1 <= d &&
              d <= day[m]) {
            record[y - 2013][m][d]++;
            if (record[y - 2013][m][d] > ans) {
              ans = record[y - 2013][m][d];
              M = m;
              D = d;
              Y = y;
            }
          }
        }
        num[0] = num[1];
        num[1] = num[2];
        num[2].clear();
      } else {
        num[2].push_back(s[i] - '0');
      }
    }
    printf("%02d-%02d-%d\n", D, M, Y);
  }
  return 0;
}
