#include "transfer.h"
#include <iostream>
#include <vector>
#include <cmath>
#include<map>
#include <algorithm>
#include <iomanip>
#include <string>
#include<stack>
#include <set>
#include <queue>
#include <chrono>
#include<array>
#include<bitset>
#include<unordered_map>
#include<random>
#include<cassert>
#include<cstring>
using namespace std;
using ll = long long;
using db = double;
const float pi = 3.14159265359;
#define V vector
#define VI V<int>
#define P pair<int,int>
#define rep(i, a, b, step) for (int i = int(a); i <= int(b); i += step)
#define repl(i,a,b,step) for (int i = int(a); i >= int(b); i -= step)
#define prime(n) [](int x) { for (int i = 2; i *1ll* i <= x; ++i) if (x % i == 0) return false; return x > 1; }(n)
#define printall(container, ch) for (const auto& elem : container) { std::cout << elem << ch; } std::cout << std::endl;
#define sn << '\n'
#define ed << endl
#define sz size()
#define print cout <<
#define debug(x) cerr<< #x << " = " << x sn;
#define mpII map<int,int>
#define mine min_element
#define maxe max_element
#define all(v) begin(v), end(v)
#define txt freopen("snakes.in", "r", stdin); freopen("snakes.out", "w", stdout)
#define pb push_back
#define pq priority_queue
#define rev reverse
#define nuyn(a) a.erase(unique(all(a)), end(a))
#define nx next_permutation
#define pk pop_back()
#define START print "Start" sn
#define END print "End" sn
#define ff first
#define ss second
#define ts to_string
#define ub upper_bound
#define mk make_pair
#define lb lower_bound
#define testcase int t;cin>>t;while(t--)solution();
std::vector<int> get_attachment(std::vector<int> sour) {
int n = sour.sz, x = 0, cnt = 0, mx;
VI ret;
if (n == 63)mx = 6;
else mx = 8;
rep(i, 0, n - 1, 1)if (sour[i])
x ^= i;
rep(i, 0, mx - 1, 1)ret.pb((x & (1 << i)) ? 1 : 0);
cnt = count(all(ret), 1) % 2;
ret.pb(cnt);
//debug("ret");
// printall(ret, ' ');
return ret;
}
std::vector<int> retrieve(std::vector<int> data) {
int n = data.sz, x = 0, mx, xr = 0, cnt = 0;
if (n <= 100)n = 63, mx = 6;
else n = 255, mx = 8;
VI ans(n);
rep(i, n, n + mx, 1)if (data[i])
x |= (1 << (i - n));
rep(i, 0, mx - 1, 1)if (x & (1 << i))
cnt ^= 1;
rep(i, 0, n - 1, 1)ans[i] = data[i];
if (cnt == data.back()) {
rep(i, 0, n - 1, 1)if (data[i])
xr ^= i;
// debug(x);
ans[xr ^ x] ^= 1;
}
return ans;
}
