The name of this problem reveals that this problem has extraordinary difficulty.

    It's well known to all peopole of Binary Indexed Tree University (BIT) that the Plastic English Club (PEC) is the second biggest club, if not the biggest club, in BIT.

    And when it comes to PEC, we have to mention ACM, who plays a vital roles in PEC.

    ACM, which stands for, American Caligraphy Management, is organized by students of BIT who was fond of caligraphy badly and is the most active part of PEC.

    Steven Jobs is chief-elder of ACM now who is also the fonder if Pear Inc.(an enormous company that focus on electronics ) and the caligraphy he had learnt in ACM club during his university helped him a lot in the fonts design of computers. For instance, serif and sanserif, two kinds of the most common fancy font styles in the world now, was designed by Steven Jobs at the very time.

    Since American Caligraphy Management is one of the major agency in PEC, they send their brilliant club members to be engaged in International Caligraphy-Pro Contest(ICPC) every years and always cut it well.

    Now, it's high time to prepare for the ICPC this year! Steven jobs is the decision maker of ICPC, so he can determine who will patricipates in ICPC representing BIT-PEC-ACM. He discovered that there are exactly  students in BIT-PEC-ACM. He can determine how many students will participate in this year ICPC and can determine who are them.

    Any arrangement is considered legal if and only if in this arrangement, Jobs choose at least one students and at most  students to go for it.

    Now, two students of BIT-PEC-ACM, Odd and Even, are discussing a funny problem. That is, Odd, as his name indicates, wants the number of students who participate in ICPC this year is an odd number, while Even, vice versa, wants it to be an even number.

    Now, as a friend of Even, can you help it to predict whose need is more likely to be met? More formally, you are supposed to calculate the value of