#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<typename K> using hset = gp_hash_table<K, null_type>;
template<typename K, typename V> using hmap = gp_hash_table<K, V>;
using namespace std;
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define eb emplace_back
#define smax(x, y) (x = max(x, y))
#define smin(x, y) (x = min(x, y))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define F0R(i, a) FOR(i,0,a)
#define ROF(i, a, b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i, a) ROF(i,0,a)
using ll = long long;
using ld = long double;
template<typename T>
using vv = vector<vector<T>>;
using vi = vector<int>;
using ii = array<int, 2>;
using vii = vector<ii>;
using vvi = vv<int>;
using vll = vector<ll>;
using l2 = array<ll, 2>;
using vl2 = vector<l2>;
using vvll = vv<ll>;
template<typename T>
using minq = priority_queue<T, vector<T>, greater<T>>;
template<typename T>
using maxq = priority_queue<T>;
template<typename T>
using oset = tree<T, null_type, less<>, rb_tree_tag, tree_order_statistics_node_update>;
//const ll M = 1000000007;
//const ll infll = M * M;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
vi a(n), k(n);
F0R(i, n) cin >> a[i];
F0R(i, n) cin >> k[i];
int B = 10;
vi c(1 << B);
F0R(i, 1 << B) c[i] = __builtin_popcount(i);
// the dp for subtask 3 is dp[x] = length of lbs ending with x
// dp[a[i]] = min(dp[x]), x & a[i] == k[i]
// however this is O(n * max(a))
// another dp is
// let dp[x][k] be the index of the lbs s.t. a[dp[x][k]] & x == k
// l[i] = dp[a[i]][k[i]] + 1
// if (a[i] & x == K) smin(dp[x][K], l[i]) (for all x, K)
// also O(n * max(a))
// but we can combine the two approaches
vi l(n, 1);
// let l[i] be the length of the lbs ending at i
vv<vi> dp(1 << B, vvi(1 << B, vi(B + 1, -1)));
// let dp[p][s][K] be the index of the lbs s.t. the last element of the lbs
// has the first B bits equal to p, and the remaining B bits (call them r)
// are such that r & s == K
// let's iterate over the high bits (p) (we are processing element at index i),
//
// now iterate over the low bits (s)
// dp[hi(a[i])][s][
int S = (1 << B) - 1;
vi rec(n, -1);
F0R(i, n) {
int p = a[i] >> B;
int s = a[i] & S;
// in this loop we want to update l[i], so we are looking at
// past values of the dp and "joining" them with a[i]
F0R(m, 1 << B) {
int K = k[i] - c[p & m]; // we already get c[p&m] bits from the prefix
if (!(0 <= K && K <= B)) continue; // this prefix is invalid
int j = dp[m][s][K];
if (j == -1) continue;
if (l[i] < l[j] + 1) {
l[i] = l[j] + 1;
rec[i] = j;
}
}
F0R(m, 1 << B) {
int &j =dp[p][m][c[s & m]];
if (j == -1 || l[j] < l[i]) {
j = i;
}
}
}
int end = max_element(all(l)) - l.begin();
cout << l[end] << '\n';
vi res;
for (; end != -1; end = rec[end]) res.pb(end);
R0F(i, res.size()) cout << res[i] + 1 << ' ';
}
