An inversion in an array is a pair of indices (i, j) such that A[i] > A[j] and i < j, but you are given two arrays A and B and  pairs such that A[i] > B[j] and i < j is called Inversion.

If total number of inversions T, is a prime number then the that will be called a MAGIC INVERSION, otherwise it will be a SIMPLE INVERSION.

INPUT

First line contains n denoting the total number of elements, N. The next line contains N space separated integers of array A.. Next line is again N space separated integers of array B.

OUTPUT

Print "MAGIC INVERSION", if total number of inversions T is prime otherwise print "SIMPLE INVERSION".

Constraints:

1 ≤ N ≤ 200000

1 ≤ A[i] ≤ 1000000

1 ≤ B[i] ≤ 1000000

SAMPLE INPUT
3
5 6 7
1 2 3

SAMPLE OUTPUT
MAGIC INVERSION

Explanation

The total number of inversions are 3, which is a prime number.
from sys import stdin
import bisect
def primes(n=10000000):
    sieve = [True] * n
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
    return [2] + [i for i in xrange(3,n,2) if sieve[i]]
def count_inversion(lst1, lst2, lst3):
    lst3.sort()
    count=0
    for i in xrange(0,len(lst1)):
        ind=bisect.bisect_left(lst3, lst1[i])
        count+=(ind-(i+1))
    return count 
prime=primes()
a=int(stdin.readline())
b=map(int, stdin.readline().split())
c=map(int, stdin.readline().split())

print 'MAGIC INVERSION' if count_inversion(b, c, c) in prime else 'SIMPLE INVERSION'