Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing *one or more stones* from any single pile. Play ends when all the stones have been removed, at which point the last player to have moved is declared the winner. Given a position in Nim, your task is to determine how many winning moves there are in that position.A position in Nim is called “losing” if the first player to move from that position would lose if both sides played perfectly. A “winning move,” then, is a move that leaves the game in a losing position. There is a famous theorem that classifies all losing positions. Suppose a Nim position contains *n* piles having *k*_{1}, *k*_{2}, …, *k*_{n} stones respectively; in such a position, there are *k*_{1} + *k*_{2} + … + *k*_{n} possible moves. We write each *k*_{i} in binary (base 2). Then, the Nim position is losing if and only if, among all the *k*_{i}’s, there are an even number of 1’s in each digit position. In other words, the Nim position is losing if and only if the *xor* of the *k*_{i}’s is 0.Consider the position with three piles given by *k*_{1} = 7, *k*_{2} = 11, and *k*_{3} = 13. In binary, these values are as follows:

111 1011 1101

The input test file will contain multiple test cases, each of which begins with a line indicating the number of piles, 1 ≤ *n* ≤ 1000. On the next line, there are n positive integers, 1 ≤ *k*_{i} ≤ 1, 000, 000, 000, indicating the number of stones in each pile. The end-of-file is marked by a test case with *n* = 0 and should not be processed.

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