#include<bits/stdc++.h>
using namespace std;
int n;
int a[100005];
int k[100005];
int prec[100005];
int dp[100005];
int ult[1030][1030][22];
signed main()
{
ios_base::sync_with_stdio(0);cin.tie(0);
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
int mxm=0,unde=-1;
for(int i=1;i<=n;i++)
{
cin>>k[i];
dp[i] = 1;
prec[i] = -1;
int curle = (a[i]&((1<<10)-1));
int curri = ((a[i]-curle)>>10);
for(int ule=0;ule<(1<<10);ule++)
{
int r = k[i] - __builtin_popcount(ule&curle);
if(dp[i] < dp[ult[ule][curri][r]]+1)
{
prec[i] = ult[ule][curri][r];
dp[i] = dp[prec[i]]+1;
}
}
for(int nri=0;nri<(1<<10);nri++)
{
if(dp[i] > dp[ult[curle][nri][__builtin_popcount(nri&curri)]])
{
ult[curle][nri][__builtin_popcount(nri&curri)] = i;
}
}
if(dp[i]>mxm)
{
mxm=dp[i];
unde=i;
}
}
vector<int> sol;
while(unde!=-1)
{
sol.push_back(unde);
unde = prec[unde];
}
cout<<sol.size()<<"\n";
for(int i=(int)sol.size()-1;i>=0;i--)
cout<<sol[i]<<" ";
return 0;
}
/**
4
1 2 3 4
10 0 1 0
2
8 9
20 0
5
5 3 5 3 5
10 1 20 1 20
ult[maskle][maskri][cnt] = lungimea maxima a unei secvente care se termina cu un element care are partea din stanga egala cu maskle si caruia,
daca ii combinam partea dreapta cu maskri, o sa avem cnt pozitii
*/
