#include <bits/stdc++.h>
#define int long long
#define x first
#define y second
#define pi acos(-1)
using namespace std;
using LL = long long;
using ULL = unsigned long long;
typedef pair<int, int> PII;
const int N = 1e6 + 10, M = 5010, mod = 1e9 + 7;
void solve() {
	int n;
	cin >> n;
	vector <int> a(n + 1);
	int p = 0;
	unordered_map <int, int> mp;
	int p2 = 0, p3 = 0;
	for(int i = 1; i <= n; i++){
		cin >> a[i];
		if(a[i] == 0){
			p = i;
		}
		if(mp[a[i]] != 0){
			p2 = mp[a[i]];
			p3 = i;
		}
		mp[a[i]] = i;
	}
	if(p != 0){
		cout << 1 << "\n";
		cout << p << "\n";
		return;
	}
	if(p2 != 0){
		cout << 2 << "\n";
		cout << p2 << " " << p3 << "\n";
		return;
	}
	/*
		使B集合里的数的各个bit位的数量都为偶数 
		考虑bit位数量大于等于2的位置 
		
		反向思考 如果某个bit位的数量为1 则相应的这个数一定使不会取的 
		迭代删除不符合的数
		最后剩下的一定能够构造出符合答案的集合 ? 奇数的情况 ? 
		7 3 6
		1 2 4
		1 2
		  2 4
		奇数定! 
		
		线性基 
	*/
	vector <PII> d(61);
	auto add = [&](int x, int y)->void{
		for(int i = 59; i >= 0; i--){
			if((x >> i) & 1){
				if(d[i].x) x ^= d[i].x;
				else{
					d[i].x = x;
					d[i].y = y;
					break;
				} 
			}
		}
		return;	
	};
	for(int i = 1; i <= n; i++){
		add(a[i], i);
	}
	int cnt = 0;
	for(int i = 0; i < 60; i++){
		if(d[i].x != 0) cnt++;
	}
	if(cnt == n || cnt == 1){
		cout << -1 << "\n";
	}else{
		cout << cnt + 1 << "\n";
		int res = 0;
		for(int i = 0; i < 60; i++){
			if(d[i].x) cout << d[i].y << " ", res ^= d[i].x;
		}
		for(int i = 1; i <= n; i++){
			if(a[i] == res) cout << i << "\n";
		}
	}
	
}
signed main() {
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int T = 1;
	//cin >> T;
	while(T--) {
		solve();
	}
}