Ancient Prophesy

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

Solution:

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>

using namespace std;

const int kd[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int n, i, j;
char s[555555];
int cnt[42][42][42];

int main() {
//  freopen("in", "r", stdin);
//  freopen("out", "w", stdout);
scanf("%s", s);
int n = strlen(s);
int mx = 0, ad = 0, am = 0, ay = 0;
memset(cnt, 0, sizeof(cnt));
for (i=0;i+9<n;i++) {
    if (s[i+2] != '-' || s[i+5] != '-') continue;
    int ok = 1;
    for (j=0;j<10;j++)
      if (j != 2 && j != 5 && s[i+j] == '-') ok = 0;
    if (!ok) continue;
    int dd = (s[i]-48)*10+(s[i+1]-48);
    int mm = (s[i+3]-48)*10+(s[i+4]-48);
    if (s[i+6] != '2' || s[i+7] != '0' || s[i+8] != '1') continue;
    int yy = s[i+9]-48;
    if (yy < 3 || yy > 5) continue;
if (mm < 1 || mm > 12) continue;
if (dd < 1 || dd > kd[mm]) continue;
cnt[dd][mm][yy]++;
if (cnt[dd][mm][yy] > mx) {
mx = cnt[dd][mm][yy];
ad = dd, am = mm, ay = yy;
}
}
printf("%d%d-%d%d-%d\n", ad/10, ad%10, am/10, am%10, 2010+ay);
return 0;
}