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30 days
has September,
April, June and November
All the rest have 31
And February’s great with 28
And Leap Year’s February’s fine with 29
The Gregorian calendar, the current standard
calendar in most part of the world, adds a 29th day to February
in all years evenly divisible by 4, except for centennial years (those ending
in -00) which are not evenly divisible by 400. Thus 1600, 2000 and 2400 are
leap years but 1700, 1800, 1900, 2100, 2200 and 2300 are not.
In
this problem, we are concerned with dates. You will be given a date and an
integer K.
You have to find the date in the calendar after K days from the given date.
Input
The first line
of input is an integer T(T<50) that represents
the number of test cases. Each case contains two lines. The first line is a
date in the format yyyy-month-dd.
year
is an integer in the range [1900, 3000], month is a string from the set {January,
February, March, April, May, June, July, August, September, October, November
and December} and dd is an integer
in the range [01,31]. The second line contains an integer K(0<K<10000).
The input date
will be a valid one.
Output
For
each input, output the case number followed by the date after K days in the
same format as that of input. Look at the sample for exact format.
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Sample Input |
Output for Sample
Input |
|
2 1984-December-30 2 1984-October-12 318 |
Case 1: 1985-January-01 Case 2:
1985-August-26 |
ProblemSetter:
Sohel Hafiz