#include "ancient2.h"
#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
typedef long long ll;
#define rep(a, b) for(int a = 0; a < (b); ++a)
#define st first
#define nd second
#define pb push_back
#define all(a) a.begin(), a.end()
bitset<1001>B[1000];
bool dodaj(bitset<1001>T) {
rep(i, 1000) if(T[i]) {
if(!B[i][i]) {
B[i]=T;
return true;
}
T^=B[i];
}
return false;
}
char znajdz(int x) {
bitset<1001>T;
T[x]=1;
rep(i, 1000) if(T[i]) T^=B[i];
return T[1000]+'0';
}
string Solve(int n) {
vector<pair<int,int>>baza;
for(int i=1; baza.size()<1000; ++i) {
rep(j, i) {
bitset<1001>T;
for(int p=j; p<1000; p+=i) T[p]=1;
if(dodaj(T)) baza.pb({i, j});
}
}
rep(i, 1000) rep(j, 1001) B[i][j]=0;
for(auto i : baza) {
vector<int>a(2*i.st), b(2*i.st);
rep(j, i.st) {
a[j]=b[j]=(j+1)%i.st;
a[j+i.st]=b[j+i.st]=(j+1)%i.st+i.st;
if(j==i.nd) {
b[j]+=i.st;
b[j+i.st]-=i.st;
}
}
bitset<1001>T;
for(int p=i.nd; p<1000; p+=i.st) T[p]=1;
if(Query(2*i.st, a, b)>=i.st) T[1000]=1;
dodaj(T);
}
string ans="";
rep(i, 1000) ans+=znajdz(i);
return ans;
}
