If n is a positive integer, define n! = n * (n-1) * (n-2) * 3 * 2 * 1, let m = 1! + 4! + +2006! + 2007!, then the sum of the last two digits of M is:

If n is a positive integer, define n! = n * (n-1) * (n-2) * 3 * 2 * 1, let m = 1! + 4! + +2006! + 2007!, then the sum of the last two digits of M is:

It's four