You are given an integer sequence 1,2,…,n1,2,…,n. You have to divide it into two sets AA and BB in such a way that each element belongs to exactly one set and |sum(A)−sum(B)||sum(A)−sum(B)| is minimum possible.
The value |x||x| is the absolute value of xx and sum(S)sum(S) is the sum of elements of the set SS.
Input
The first line of the input contains one integer nn (1≤n≤2⋅1091≤n≤2⋅109).
Output
Print one integer — the minimum possible value of |sum(A)−sum(B)||sum(A)−sum(B)| if you divide the initial sequence 1,2,…,n1,2,…,n into two sets AA and BB.
Examples
Input
3
Output
0
Input
5
Output
1
Input
6
Output
1
Note
Some (not all) possible answers to examples:
In the first example you can divide the initial sequence into sets A={1,2} and B={3} so the answer is 00.
In the second example you can divide the initial sequence into sets A={1,3,4} and B={2,5} so the answer is 11.
In the third example you can divide the initial sequence into sets A={1,4,5}and B={2,3,6} so the answer is 11.
解题思路：先求出前n项的和，如果前n项和为偶数，则总能找到两个子集的和相等；若为奇数，则总能找到两个子集和相差为1.
include<stdio.h>
int main()
{
long long n,ans;
while(~scanf("%lld",&n))
{
ans=(1+n)*n/2;
if(ans%2==1)
printf("1\n");
else if(ans%2==0)
printf("0\n");
}
}