Problem: 2^N players competed in a tournament. Each player has a unique ID number from 1 through 2^N. When two players play a match, the player with the smaller ID always wins.

This tournament was a little special, in which losers are not eliminated to fully rank all the players.

We will call such a tournament that involves 2^n players a full tournament of level n. In a full tournament of level n, the players are ranked as follows:

* In a full tournament of level 0, the only player is ranked first.
* In a full tournament of level n\ (1 \leq n), initially all 2^n players are lined up in a row. Then:
* First, starting from the two leftmost players, the players are successively divided into 2^{n-1} pairs of two players.
* Each of the pairs plays a match. The winner enters the "Won" group, and the loser enters the "Lost" group.
* The players in the "Won" group are lined up in a row, maintaining the relative order in the previous row. Then, a full tournament of level n-1 is held to fully rank these players.
* The players in the "Lost" group are also ranked in the same manner, then the rank of each of these players increases by 2^{n-1}.



For example, the figure below shows how a full tournament of level 3 progresses when eight players are lined up in the order 3,4,8,6,2,1,7,5. The list of the players sorted by the final ranking will be 1,3,5,6,2,4,7,8.

<image>

Takahashi has a sheet of paper with the list of the players sorted by the final ranking in the tournament, but some of it blurred and became unreadable. You are given the information on the sheet as a sequence A of length N. When A_i is 1 or greater, it means that the i-th ranked player had the ID A_i. If A_i is 0, it means that the ID of the i-th ranked player is lost.

Determine whether there exists a valid order in the first phase of the tournament which is consistent with the sheet. If it exists, provide one such order.

Constraints

* 1 \leq N \leq 18
* 0 \leq A_i \leq 2^N
* No integer, except 0, occurs more than once in A_i.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_{2^N}


Output

If there exists a valid order in the first phase of the tournament, print `YES` first, then in the subsequent line, print the IDs of the players sorted by the final ranking, with spaces in between. If there is no such order, print `NO` instead.

Examples

Input

3
0 3 0 6 0 0 0 8


Output

YES
3 4 8 6 2 1 7 5


Input

1
2 1


Output

NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

using ll=long long;
#define int ll

#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define pb push_back
#define eb emplace_back
#define a first
#define b second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#ifdef LOCAL
#define dmp(x) cerr<<__LINE__<<" "<<#x<<" "<<x<<endl
#else
#define dmp(x) void(0)
#endif

template<class t,class u> void chmax(t&a,u b){if(a<b)a=b;}
template<class t,class u> void chmin(t&a,u b){if(b<a)a=b;}

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class t,class u>
ostream& operator<<(ostream& os,const pair<t,u>& p){
	return os<<"{"<<p.a<<","<<p.b<<"}";
}

template<class t> ostream& operator<<(ostream& os,const vc<t>& v){
	os<<"{";
	for(auto e:v)os<<e<<",";
	return os<<"}";
}

#define mp make_pair
#define mt make_tuple
#define one(x) memset(x,-1,sizeof(x))
#define zero(x) memset(x,0,sizeof(x))
#ifdef LOCAL
void dmpr(ostream&os){os<<endl;}
template<class T,class... Args>
void dmpr(ostream&os,const T&t,const Args&... args){
	os<<t<<" ";
	dmpr(os,args...);
}
#define dmp2(...) dmpr(cerr,"Line:",__LINE__,##__VA_ARGS__)
#else
#define dmp2(...) void(0)
#endif

using uint=unsigned;
using ull=unsigned long long;

template<int i,class T>
void print_tuple(ostream&,const T&){
}

template<int i,class T,class H,class ...Args>
void print_tuple(ostream&os,const T&t){
	if(i)os<<",";
	os<<get<i>(t);
	print_tuple<i+1,T,Args...>(os,t);
}

template<class ...Args>
ostream& operator<<(ostream&os,const tuple<Args...>&t){
	os<<"{";
	print_tuple<0,tuple<Args...>,Args...>(os,t);
	return os<<"}";
}

void print(ll x,int suc=1){
	cout<<x;
	if(suc==1)
		cout<<"\n";
	if(suc==2)
		cout<<" ";
}

ll read(){
	ll i;
	cin>>i;
	return i;
}

vi readvi(int n,int off=0){
	vi v(n);
	rep(i,n)v[i]=read()+off;
	return v;
}

template<class T>
void print(const vector<T>&v,int suc=1){
	rep(i,v.size())
		print(v[i],i==int(v.size())-1?suc:2);
}

string readString(){
	string s;
	cin>>s;
	return s;
}

template<class T>
T sq(const T& t){
	return t*t;
}

#define CAPITAL
void yes(bool ex=true){
	#ifdef CAPITAL
	cout<<"YES"<<endl;
	#else
	cout<<"Yes"<<endl;
	#endif
	if(ex)exit(0);
}
void no(bool ex=true){
	#ifdef CAPITAL
	cout<<"NO"<<endl;
	#else
	cout<<"No"<<endl;
	#endif
	if(ex)exit(0);
}

constexpr ll ten(int n){
	return n==0?1:ten(n-1)*10;
}

const ll infLL=LLONG_MAX/3;

#ifdef int
const int inf=infLL;
#else
const int inf=INT_MAX/2-100;
#endif

int topbit(signed t){
	return t==0?-1:31-__builtin_clz(t);
}
int topbit(ll t){
	return t==0?-1:63-__builtin_clzll(t);
}
int botbit(signed a){
	return a==0?32:__builtin_ctz(a);
}
int botbit(ll a){
	return a==0?64:__builtin_ctzll(a);
}
int popcount(signed t){
	return __builtin_popcount(t);
}
int popcount(ll t){
	return __builtin_popcountll(t);
}
bool ispow2(int i){
	return i&&(i&-i)==i;
}
int mask(int i){
	return (int(1)<<i)-1;
}

bool inc(int a,int b,int c){
	return a<=b&&b<=c;
}

template<class t> void mkuni(vc<t>&v){
	sort(all(v));
	v.erase(unique(all(v)),v.ed);
}

ll rand_int(ll l, ll r) { //[l, r]
	#ifdef LOCAL
	static mt19937_64 gen;
	#else
    static random_device rd;
    static mt19937_64 gen(rd());
    #endif
    return uniform_int_distribution<ll>(l, r)(gen);
}

template<class t>
int lwb(const vc<t>&v,const t&a){
	return lower_bound(all(v),a)-v.bg;
}

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	cout<<fixed<<setprecision(20);
	
	int n;cin>>n;
	int s=1<<n;
	vi a=readvi(s,-1);
	vi b(s,s-1);
	per(i,s){
		if(a[i]!=-1)
			b[i]=a[i];
		rep(j,n)if(!(i&1<<j))
			chmin(b[i],b[i^1<<j]);
	}
	vi c(s,0);
	rep(i,s){
		if(a[i]!=-1)
			c[i]=a[i];
		rep(j,n)if(i&1<<j)
			chmax(c[i],c[i^1<<j]);
	}
	dmp(b);
	dmp(c);
	vi d(s);
	vc<tuple<int,int,int>> qs;
	rep(i,s)
		qs.eb(b[i],c[i],i);
	sort(all(qs));
	set<int> rem;
	rep(i,s+1)rem.insert(i);
	for(auto q:qs){
		int l,r,i;tie(r,l,i)=q;
		dmp(q);
		auto itr=rem.lower_bound(l);
		if(*itr>r)no();
		d[i]=*itr;
		rem.erase(itr);
	}
	rep(i,s){
		int j=0;
		rep(k,n)if(i&1<<k)j+=1<<(n-1-k);
		if(i<j)swap(d[i],d[j]);
	}
	for(auto&v:d)v++;
	yes(0);
	print(d);
}