Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) A

Description

You are given names of two days of the week.

Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year.

In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.

Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".

Input

The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".

Output

Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes).

Examples

input

monday
tuesday

output

NO

input

sunday
sunday

output

YES

input

saturday
tuesday

output

YES

Note

In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays.

In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday.

题意：给我两个日期表示星期几，现在问，如果一个月的一号是前一个数，那么下一个月的一号是不是后面一个数字

解法：不计闰年，那么把这一个月所有可能的天数都加一下%7，看余数是不是后面一个数字

#include <bits/stdc++.h>
using namespace std;
int main(){
    string s1,s2;
    map<string,int>v;
    v["monday"]=1;
    v["tuesday"]=2;
    v["wednesday"]=3;
    v["thursday"]=4;
    v["friday"]=5;
    v["saturday"]=6;
    v["sunday"]=0;
    cin>>s1>>s2;
    int flag=0,a=v[s1],b=v[s2];
    if((a+28)%7==b || (a+30)%7==b || (a+31)%7==b) flag=1;
    if(flag) cout<<"YES\n";
    else cout<<"NO\n";
    return 0;
}