#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
int read(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
const int N = 550000;
const int Mod = 998244353;
int inv = 499122177;
int V = (1<<19);
void fwt(int *a, int z) {
  	for(int mid=1;mid<V;mid<<=1){
  		int k = (mid<<1);
	    for(int i=0;i<V;i+=k){
	      	for(int j=0;j<mid;j++){
	        	int x = a[i+j],y=a[i+j+mid];
	        	a[i+j] = 1ll*z*(x+y)%Mod;
	        	a[i+j+mid] = (1ll*z*(x-y+Mod)%Mod+Mod)%Mod;
	      	}
	    }
  	}
}
int a[20][N];
int n;
int main(){
	n = read();
	int sum = 0;
	for(int i=1;i<=n;i++){
		int x = read();
		a[1][x] = 1;
		sum^=x;
	}
	if(sum==0){
		cout<<n<<"\n";
		return 0;
	}
	a[1][0] = 1;
	fwt(a[1],1);
	for(int i=2;i<=19;i++){
		for(int j=0;j<V;j++){
			a[i][j] = 1ll*a[i-1][j]*a[1][j]%Mod;
		}
	}
	int l = 1,r = 19;
	int ans = 0;
	while(l<=r){
		int mid = (l+r)/2;
		fwt(a[mid],inv);
		if(a[mid][sum]){
			ans = mid;
			r = mid-1;
		}else{
			l = mid+1;
		}
	}
	cout<<n-ans<<"\n";
	return 0;
}
/*
直接卷积表示删i个数好像就做完了，赢
最多删Log个数 
fwt真是太美丽了
俩log能过吗 
哦可以二分，变loglog
能过吗/jk 
*/