The Universidad Nacional de Colombia (UN) was created on September 22 of 1867, a Sunday. In 2017 the UN had its sesquicentenary, i.e. it was its 150th birthday. The sesquicentenary of the UN was on a Friday. As it is really important to know all the details corresponding to the date of this huge celebration for every year, we want to know which day of the week corresponds to a specific birthday of the Universidad Nacional de Colombia.

Remember that a year usually has 365 days, except for some years, called leaps-years which have 366 days. To know which year is a leap-year we have the following rule: Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100, but these centurial years are leap years if they are exactly divisible by 400. For example, the years 1700, 1800, and 1900 were not leap years, but the years 1600 and 2000 were.

Input

The input consists of a number n (0 ≤ n ≤ 10000)

Output

Output just one word corresponding to the day of the week at the n - th birthday of the UN.

Print the day on the following format (without quotation marks): 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday' or 'Sunday'.

Examples

Input

0

Output

Sunday

Input

150

Output

Friday

題意：在1867年九月份星期天生日，問你，給出第N個生日給你，問你是星期幾生日；

思路：若是閏年，則在下一年加二，否加一；

具體看程式碼：

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
int main()
{

    int n;
    scanf("%d",&n);
    int p=1867;
    int ans=6;
    for(int i=p; i<=n+p; i++)
    {
        if((i%4==0&&i%100!=0)||(i%400==0))
            ans+=2;
        else ans+=1;
    }
    int t;
    t=ans%7;
    if(t==1)
        printf("Monday\n");
    else if(t==2)
        printf("Tuesday\n");
    else if(t==3)
        printf("Wednesday\n");
    else if(t==4)
        printf("Thursday\n");
    else if(t==5)
        printf("Friday\n");
    else if(t==6)
        printf("Saturday\n");
    else if(t==0)
        printf("Sunday\n");
    return 0;
}