#include "sphinx.h"
#include <bits/stdc++.h>
using namespace std;
#define vi vector<int>
#define pb push_back
#define pii pair<int,int>
#define ff first
#define ss second
#define all(x) x.begin(),x.end()
#define ll long long int
const int N = 250;
vi g[N];
int ask(vi E){
return perform_experiment(E);
}
std::vector<int> find_colours(int n, std::vector<int> X, std::vector<int> Y) {
for(int i = 0; i < n; ++i) g[i].clear();
for(int i = 0; i < n - 1; ++i) g[X[i]].pb(Y[i]), g[Y[i]].pb(X[i]);
vi col(n, -1);
for(int j = 0; j < n; ++j){
vi E(n, j);
for(int i = 0; i < n; i += 2){
E[i] = -1;
}
if(ask(E) < n){
vi S;
for(int i = 0; i < n; i += 2) S.pb(i);
// that means there is someone with color j
bool ok = true;
int L = 0;
while(L < int(S.size())){
ok = false;
int l = L, r = int(S.size()) - 1, res = int(S.size());
while(l <= r){
int mid = l+r>>1;
vi EE(n, j);
int cnt = 0, is = 0;
for(int i = L; i <= mid; ++i) EE[S[i]] = -1, ++cnt;
if(S[L] > 0) ++is;
if(S[mid] < n - 1) ++is;
if(ask(EE) < cnt * 2 - 1 + is){
res = mid;
r = mid - 1;
}else{
l = mid + 1;
}
}
if(res == int(S.size())) break;
col[S[res]] = j;
L = res + 1;
if(L >= int(S.size())) break;
vi EE(n, j);
int cnt = 0, is = 0;
for(int i = L; i < int(S.size()); ++i) EE[S[i]] = -1, ++cnt;
if(S[L] > 0) ++is;
if(S.back() < n - 1) ++is;
if(ask(EE) == cnt * 2 - 1 + is){
break;
}
}
}
}
for(int j = 0; j < n; ++j){
vi E(n, j);
for(int i = 1; i < n; i += 2){
E[i] = -1;
}
if(ask(E) < n){
vi S;
for(int i = 1; i < n; i += 2) S.pb(i);
// that means there is someone with color j
bool ok = true;
int L = 0;
while(L < int(S.size())){
ok = false;
int l = L, r = int(S.size()) - 1, res = int(S.size());
while(l <= r){
int mid = l+r>>1;
vi EE(n, j);
int cnt = 0, is = 0;
for(int i = L; i <= mid; ++i) EE[S[i]] = -1, ++cnt;
if(S[L] > 0) ++is;
if(S[mid] < n - 1) ++is;
if(ask(EE) < cnt * 2 - 1 + is){
res = mid;
r = mid - 1;
}else{
l = mid + 1;
}
}
if(res == int(S.size())) break;
col[S[res]] = j;
L = res + 1;
if(L >= int(S.size())) break;
vi EE(n, j);
int cnt = 0, is = 0;
for(int i = L; i < int(S.size()); ++i) EE[S[i]] = -1, ++cnt;
if(S[L] > 0) ++is;
if(S.back() < n - 1) ++is;
if(ask(EE) == cnt * 2 - 1 + is){
break;
}
}
}
}
return col;
}
