#include <bits/stdc++.h>
using namespace std;

#define QwQ330AwA return 0
#define ll long long

const int mod = 998244353;
int n, m;
int f[1 << 19];
int g[1 << 19];

void FWT(int *f, int op) {
	for (int mid = 1, len = 2; len <= (1 << m); mid <<= 1, len <<= 1) {
		for (int i = 0; i < (1 << m); i += len) {
			for (int j = 0; j < mid; j++) {
				int x = f[i + j], y = f[i + j + mid];
				f[i + j] = x + y;
				if (f[i + j] >= mod) f[i + j] -= mod;
				f[i + j + mid] = x - y;
				if (f[i + j + mid] < 0) f[i + j + mid] += mod;
			}
		}
	}
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	int sum = 0;
	for (int i = 1, x; i <= n; i++) {
		cin >> x;
		sum ^= x;
		f[x] = 1;
	}
	if (sum == 0) {
		cout << n;
		return 0;
	}
	m = 19;
	FWT(f, 1);
	for (int s = 0; s < (1 << m); s++) g[s] = 1;
	for (int i = 1; ; i++) {
		int fwt = 0;
		for (int s = 0; s < (1 << m); s++) {
			g[s] = 1ll * g[s] * f[s] % mod;
			if (__builtin_popcount(s & sum) & 1) fwt = (fwt - g[s] + mod) % mod;
			else fwt = (fwt + g[s]) % mod;
		}
		if (fwt) {
			cout << n - i;
			return 0;
		}
	}
	QwQ330AwA;
}
/*
  就是选一些数等于异或值，最小化选的个数
  答案不会很大，线性基最多只有logV
  就是异或意义下做卷积，枚举答案判断是否有那一位就可以
  O(2^nlognlogV)?
  哦单点求值用定义就好了
  O(2^nlogn)
 */