#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
const int N = 1e5;
const int B = 10;
int bc[1 << B][1 << B]; // bc[i][j] = bit_count(i & j)
struct State {
int len;
int end;
} dp[1 << B][1 << B][B + 1];
int main() {
// preprocess:
for (int i = 0; i < (1 << B); i++) {
for (int j = 0; j < (1 << B); j++) { bc[i][j] = __builtin_popcount(i & j); }
}
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
vector<int> k(n);
for (int i = 0; i < n; i++) cin >> k[i];
int ans = 1;
// best_i: most optimal index to end on
int best_i = 0;
vector<int> prv(n); // prv[i]: best index to include before i
iota(prv.begin(), prv.end(), 0); // prv[i] = i
for (int i = 0; i < n; i++) {
int l = a[i] >> B; // l(a[i])
int r = a[i] % (1 << B); // r(a[i])
int lbs = 1; // length of longest lbs that ends at i
// enumerate all possibilities for l(prev_num)
for (int j = 0; j < (1 << B); j++) {
// here, we use the fact that
// bc[x][y] = bc[l(x)][l(y)] + bc[r(x)][r(y)],
// or bc[r(x)][r(y)] = bc[x][y] - bc[l(x)][l(y)]
// required value of bc[r(x)][r(y)]
int needed = k[i] - bc[j][l];
if (needed < 0 || needed > B) continue;
if (dp[j][r][needed].len + 1 > lbs) {
lbs = dp[j][r][needed].len + 1;
prv[i] = dp[j][r][needed].end;
}
}
if (lbs > ans) {
ans = lbs;
best_i = i;
}
// update all answers a[i] affects
for (int j = 0; j < (1 << B); j++) {
State &new_state = dp[l][j][bc[r][j]];
if (lbs > new_state.len) {
new_state.len = lbs;
new_state.end = i;
}
}
}
cout << ans << endl;
vector<int> res;
while (prv[best_i] != best_i) {
res.push_back(best_i);
best_i = prv[best_i];
}
res.push_back(best_i);
reverse(res.begin(), res.end());
for (int x : res) { cout << x + 1 << " "; }
}
