> Even or Odd
> 
> Time Limit : 5 seconds Memory Limit : 10MB
> 
> We define f(p1,p2,…,pn) as the number of inversion pairs(an inversion pair is such a pair (pi, pj) which i < j and pi > pj). Here p1p2p3…pn is a permutation of 123..n, i.e. {p1,p2,p3,…,pn} = {1,2,…, n}.
> 
> We call permutation p1p2p3…pn an even permutation if f(p1,p2,…,pn) is even. As you can guess, the other permutations are called odd permutations.
> 
> Maybe you can easily calculate the function f, but we are only interested in the value (-1) f(p1,p2,…,pn).For this purpose, we want to know whether a permutation is even.
> 
> For example f(1,2,3,4) = 0, so 1234 is a even permutation; f(2,4,1,3) = 3, thus 2413 is an odd permutation.
> 
> Input:
> 
> The input contains several cases.
> 
> Each case has two lines, an integer N (1<=N<=2000000) on the first line and the permutation on the second line(numbers are separated with a blank). Input ends with an zero in a single line.
> 
> Output:
> 
> For each case, output one line. Output “EVEN” if the permutation is an even permutation, otherwise output “ODD”.
> 
> Sample Input:
> 
> 2
> 1 2
> 4
> 2 4 1 3
> 0
> 
> Sample Output:
> 
> EVEN
> ODD
> 
> Hint:
> 
> Huge input, cin(c++) and Scanner(Java) will TLE!