/*

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 k1, k2, …, kn stones respectively;

in such a position, there are k1 + k2 + … + kn possible

moves. We write each ki in binary (base 2). Then, the Nim

position is losing if and only if, among all the ki’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 ki’s is 0.

Consider the position with three piles given by k1 = 7, k2

 = 11, and k3 = 13. In binary, these values are as follows:

                           111

                          1011

                          1101

There are an odd number of 1’s among the rightmost digits,

so this position is not losing. However, suppose k3 were

changed to be 12. Then, there would be exactly two 1’s in

each digit position, and thus, the Nim position would become

losing. Since a winning move is any move that leaves the

game in a losing position, it follows that removing one

stone from the third pile is a winning move when k1 = 7, k2

= 11, and k3 = 13. In fact, there are exactly three winning

moves from this position: namely removing one stone from any

of the three piles.

*/

#include <set>

#include <map>

#include <cmath>

#include <queue>

#include <vector>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

int n,k[];

int main(){

    while(scanf("%d",&n)){

        if(n==)break;

        int sg=,ans=;

        for(int i=;i<=n;i++)scanf("%d",&k[i]),sg^=k[i];

        for(int i=;i<=n;i++)if((k[i]^sg)<k[i])ans++;

        printf("%d\n",ans);

    }

    return ;

}