Given is an integer N. Find the minimum possible positive integer k such that (1+2+\cdots+k) is a multiple of N. It can be proved that such a positive integer k always exists.

Constraints

* 1 \leq N \leq 10^{15}
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N


Output

Print the answer in a line.

Examples

Input

11


Output

10


Input

20200920


Output

1100144
def inv_gcd(a,b):
    a=a%b
    if a==0:
        return (b,0)
    s=b;t=a
    m0=0;m1=1
    while(t):
        u=s//t
        s-=t*u
        m0-=m1*u
        s,t=t,s
        m0,m1=m1,m0
    if m0<0:
        m0+=b//s
    return (s,m0)
def inv_mod(x,m):
    assert 1<=m
    z=inv_gcd(x,m)
    assert z[0]==1
    return z[1]
def crt(r,m):
    assert len(r)==len(m)
    n=len(r)
    r0=0;m0=1
    for i in range(n):
        assert 1<=m[i]
        r1=r[i]%m[i]
        m1=m[i]
        if m0<m1:
            r0,r1=r1,r0
            m0,m1=m1,m0
        if (m0%m1==0):
            if (r0%m1!=r1):
                return (0,0)
            continue
        g,im=inv_gcd(m0,m1)
        u1=m1//g
        if ((r1-r0)%g):
            return (0,0)
        x=(r1-r0)//g % u1*im%u1
        r0+=x*m0
        m0*=u1
        if r0<0:
            r0+=m0
    return (r0,m0)
N=int(input())
def divide(n):
    res=[]
    i=1
    while(i*i<=n):
        if n%i==0:
            res.append(i)
            if i*i!=n:
                res.append(n//i)
        i+=1
    return res
ans=2*N
for b in divide(2*N):
    c=(2*N)//b
    tmp=crt([0,c-1],[b,c])
    if tmp==(0,0):
        continue
    x=tmp[0]
    if x==0:
        x=tmp[1]
    ans=min(x,ans)
print(ans)