#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using db = long double;
#define ff first
#define ss second
#define pb push_back
#define all(x) begin(x),end(x)
#define FOR(i,a,b) for(int i = (a);i<(b);i++)
#define F0R(i,a) FOR(i,0,a)
#define rep(a) F0R(_,a)
#define each(a,x) for(auto& a:x)
#define sz(x) (int)size(x)
#define vi vector<int>
const int MOD = 1e9+7;
const db PI = acos((db)-1);
int main(){
cin.tie(0)->sync_with_stdio(0);
int n; cin>>n;
vi A(n+1), K(n+1);
int mx = -1;
FOR(i,1,n+1){
cin>>A[i];
mx = max(mx,A[i]);
}
FOR(i,1,n+1){
cin>>K[i];
}
if(n<=5000){
vi dp(n+1,1), memo(n+1,-1);
FOR(i,1,n+1){
FOR(j,1,i){
if(__builtin_popcount(A[i]&A[j]) == K[i] && dp[j]+1>dp[i]){
dp[i] = dp[j] + 1;
memo[i] = j;
}
}
}
int ans = 0, id = -1;
FOR(i,1,n+1){
if(dp[i]>ans){
id = i;
ans = dp[i];
}
}
cout<<ans<<"\n";
vi sol;
while(id != -1){
sol.pb(id);
id = memo[id];
}
reverse(all(sol));
each(x,sol){
cout<<x<<" ";
}
}else if(mx < 256){
vi dp(mx+1,-1),memo(mx+1,-1),ant(n+1,-1);
FOR(i,1,n+1){
dp[A[i]] = 1;
if(memo[A[i]] == -1) memo[A[i]] = i;
}
// dp(x) é a maior subsequência que termina no valor x
FOR(i,1,n+1){
FOR(j,0,mx+1)if(dp[j] != -1 && __builtin_popcount(A[i] & j) == K[i]){
if(dp[j]+1 > dp[A[i]]){
dp[A[i]] = dp[j]+1;
memo[A[i]] = i;
ant[i] = memo[j];
}
}
}
int ans = 0, val = -1;
FOR(i,0,mx+1){
if(ans<dp[i]){
val = i;
ans = dp[i];
}
}
cout<<ans<<"\n";
val = memo[val];
vi sol;
while(val != -1){
sol.pb(val);
val = ant[val];
}
reverse(all(sol));
each(x,sol){
cout<<x<<" ";
}
}
return 0;
}
