【codeforces】New Year and Days

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一只名叫Limak的小北极熊决定在2016年全年定期储蓄糖果，并选择了一个特定的计划来执行这一任务。该计划可以是在每周的固定某一天或是每月的固定某一天进行储蓄。根据所选的日期，需要计算出在2016年里能够储蓄的糖果总数。

New Year and Days

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.

Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.

Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.

Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.

Input

The only line of the input is in one of the following two formats:

"x of week" where x (1 ≤ x ≤ 7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday.
"x of month" where x (1 ≤ x ≤ 31) denotes the day of the month.

Output

Print one integer — the number of candies Limak will save in the year 2016.

Sample Input

Input

4 of week一年中有几个星期四

Output

Input

30 of month一年中有几个每月的第30天

Output

Hint

Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – https://en.wikipedia.org/wiki/Gregorian_calendar. The week starts with Monday.

In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.

In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.

#include<stdio.h>
#include<string.h>
int main()
{
	int n;
	char a[5],b[10];
	while(scanf("%d%s%s",&n,&a,&b)!=EOF)
	{
		int t;
		if(b[0]=='w')
		{
			if(n==5||n==6)
			t=366/7+1;
			else
			t=366/7;
			printf("%d\n",t);
		}
		else
		{
			if(n<=29)
			t=12;
			else if(n==30)
			t=11;
			else 
			t=7;
			
			printf("%d\n",t);
		}
	}
	
	return 0;
}