#include <bits/stdc++.h>
using namespace std;
void testCase() {
int n;
cin >> n;
vector<int> a(n), k(n);
for (int &x : a) {
cin >> x;
}
for (int &x : k) {
cin >> x;
}
/// len[i] =def= cel mai lung subsir care satisface proprietatile din enunt si se termina la pozitia i
/// dp[trueMask][otherMask][set] =def= valoarea maxima len[i] de pana acum astfel incat a[i]
/// are prima jumatate firstHalf = trueMask si count(secondHalf & otherMask) = set
/// Astfel, cand vreau sa aflu len[i], pot fixa prima jumatate a mastii de dinainte,
/// fie aceasta mask si ma intereseaza atunci pentru tranzitie dp[mask][secondHalf][need],
/// unde need = k[i] - count(firstHalf & mask)
/// Complexitate: O(N * sqrt(MAX_VAL))
const int m = (1 << 10);
const int lg = 10;
vector<int> len(n, 1), last(n, -1), cnt(m);
vector<vector<vector<int>>> dp(m, vector<vector<int>>(m, vector<int>(1 + lg, -1)));
for (int i = 1; i < m; ++i) {
cnt[i] = __builtin_popcount(i);
}
for (int i = 0; i < n; ++i) {
int firstHalf = 0, secondHalf = 0;
for (int bit = 0; (1 << bit) <= a[i]; ++bit) {
if (a[i] & (1 << bit)) {
if (bit < lg) {
firstHalf |= (1 << bit);
} else {
secondHalf |= (1 << (bit - 10));
}
}
}
for (int mask = 0; mask < m; ++mask) {
int need = k[i] - cnt[mask & firstHalf];
if (0 <= need && need <= lg) {
int index = dp[mask][secondHalf][need];
if (index != -1 && len[i] < len[index] + 1) {
len[i] = len[index] + 1;
last[i] = index;
}
}
}
for (int mask = 0; mask < m; ++mask) {
int common = cnt[secondHalf & mask];
int index = dp[firstHalf][mask][common];
if (index == -1 || len[index] < len[i]) {
dp[firstHalf][mask][common] = i;
}
}
}
int pos = 0;
for (int i = 1; i < n; ++i) {
if (len[pos] < len[i]) {
pos = i;
}
}
vector<int> sol;
while (pos != -1) {
sol.emplace_back(pos + 1);
pos = last[pos];
}
reverse(sol.begin(), sol.end());
cout << sol.size() << '\n';
for (int x : sol) {
cout << x << ' ';
}
cout << '\n';
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int tests = 1;
for (int tc = 0; tc < tests; ++tc) {
testCase();
}
return 0;
}
