#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()
#define fi first
#define se second
template<class T> bool ckmin(T &a, const T &b) {return a > b ? a = b, 1 : 0;}
template<class T> bool ckmax(T &a, const T &b) {return a < b ? a = b, 1 : 0;}
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for(auto z : x) cerr << (f++ ? "," : ""), __print(z); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if(sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef ljuba
#define dbg(x...) cerr << "LINE(" << __LINE__ << ") -> " << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif
const char nl = '\n';
#include "supertrees.h"
const int mxN = 1005;
bool vis[mxN];
int n;
vi cur;
vvi p;
vvi ans;
struct DSU {
int rod[mxN], vel[mxN];
DSU() {
for(int i = 0; i < mxN; ++i) {
rod[i] = i;
vel[i] = 1;
}
}
int find_set(int a) {
if(a == rod[a])
return a;
return rod[a] = find_set(rod[rod[a]]);
}
bool unite(int a, int b) {
a = find_set(a);
b = find_set(b);
if(a == b)
return 0;
if(vel[a] > vel[b])
swap(a, b);
vel[b] += vel[a];
rod[a] = b;
return 1;
}
}dsu;
void dfs(int s) {
cur.pb(s);
dbg(s);
vis[s] = 1;
for(int i = 0; i < n; ++i) {
if(!vis[i] && p[s][i])
dfs(i);
}
}
bool idi(int s) {
cur.clear();
dfs(s);
for(auto z : cur) {
for(auto z2 : cur) {
if(!p[z][z2])
return 0;
}
}
//resenje bi trebalo da izgleda kao da dole imas lance, a iznad lanaca jos dva cvora tako da su oba povezana sa pocetkom svih lanaca
vi lanac;
dbg(cur);
for(auto z : cur) {
if(dsu.find_set(z) == z) {
lanac.pb(z);
vi comp;
for(int i = 0; i < n; ++i) {
if(dsu.find_set(i) == z) {
comp.pb(i);
}
}
dbg(comp);
for(int i = 0; i + 1 < sz(comp); ++i) {
ans[comp[i]][comp[i+1]] = ans[comp[i+1]][comp[i]] = 1;
}
for(auto z : comp) {
for(auto z2 : comp)
if(p[z][z2] != 1)
return 0;
}
}
}
dbg(lanac);
if(sz(lanac) == 2)
return 0;
if(sz(lanac) == 1)
return 1;
for(int i = 0; i < sz(lanac); ++i) {
ans[lanac[i]][lanac[(i+1)%sz(lanac)]] = ans[lanac[(i+1)%sz(lanac)]][lanac[i]] = 1;
}
return 1;
}
int construct(vvi p2) {
p = p2;
n = sz(p);
ans = vvi(n, vi(n, 0));
for(int i = 0; i < n; ++i) {
for(int j = 0; j < n; ++j) {
if(p[i][j] == 3)
return 0;
}
}
for(int i = 0; i < n; ++i) {
for(int j = 0; j < n; ++j) {
if(p[i][j] == 1)
dsu.unite(i, j);
}
}
for(int i = 0; i < n; ++i) {
if(!vis[i]) {
if(!idi(i))
return 0;
}
}
dbg(ans);
build(ans);
return 1;
}
