#include<bits/stdc++.h>
using namespace std;
using i64 = long long;
#define int i64
#define vi vector<int>
#define vvi vector<vi>
#define vb vector<bool>
#define pii pair<int, int>
#define fi first
#define se second
#define sz(x) (int)(x).size()
const int B = 10;
struct State{
int len, end;
};
signed main() {
ios_base::sync_with_stdio(false); cin.tie(0);
vvi bc(1 << B, vi(1 << B));
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, ans = 1, best_i = 0;
cin >> n;
vi a(n), k(n);
for(int &i : a)
cin >> i;
for(int &i : k)
cin >> i;
vi prev(n);
iota(prev.begin(), prev.end(), 0);
vector<vector<vector<State>>> dp(1 << B, vector<vector<State>>(1 << B, vector<State>(B + 1)));
for(int i = 0; i < n; i++){
int l = a[i] >> B, r = a[i] % (1 << B), lbs = 1;
for(int j = 0; j < (1 << B); j++){ // enumerate all possibilities for l(prev_num)
// 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)]
int needed = k[i] - bc[j][l]; // required value of bc[r(x)][r(y)]
if(needed < 0 || needed > B) continue;
if(dp[j][r][needed].len + 1 > lbs){
lbs = dp[j][r][needed].len + 1;
prev[i] = dp[j][r][needed].end;
}
}
if(lbs > ans){
ans = lbs;
best_i = i;
}
for(int j = 0; j < (1 << B); j++){ // update all answers a[i] affects
State &new_state = dp[l][j][bc[r][j]];
if(lbs > new_state.len){
new_state.len = lbs;
new_state.end = i;
}
}
}
cout << ans << "\n";
vi res;
while(prev[best_i] != best_i){
res.emplace_back(best_i + 1);
best_i = prev[best_i];
}
res.emplace_back(best_i + 1);
reverse(res.begin(), res.end());
for(int i : res){
cout << i << " ";
}
return 0;
}
