Problem:
Score : 1600 points

Problem Statement

You are given a string S of length 2N, containing N occurrences of a and N occurrences of b.

You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of a and the i-th occurrence of b. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).

Find the lexicographically largest string that can be obtained in this way.

Constraints

1 \leq N \leq 3000

S is a string of length 2N containing N occurrences of a and N occurrences of b.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the lexicographically largest string that satisfies the condition.

Sample Input 1

3
aababb

Sample Output 1

abab

A subsequence of T obtained from taking the first, third, fourth and sixth characters in S, satisfies the condition.

Sample Input 2

3
bbabaa

Sample Output 2

bbabaa

You can choose all the characters.

Sample Input 3

6
bbbaabbabaaa

Sample Output 3

bbbabaaa

Sample Input 4

9
abbbaababaababbaba

Sample Output 4

bbaababababa

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
	using namespace std;
	#define pii pair<int,int>
	#define fir first
	#define sec second
	#define MP make_pair
	#define rep(i,x,y) for (int i=(x);i<=(y);i++)
	#define drep(i,x,y) for (int i=(x);i>=(y);i--)
	#define go(x) for (int i=head[x];i;i=edge[i].nxt)
	#define templ template<typename T>
	#define sz 3300
	typedef long long ll;
	typedef double db;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
	templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
	templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
	templ inline void read(T& t)
	{
		t=0;char f=0,ch=getchar();double d=0.1;
		while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
		while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
		if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
		t=(f?-t:t);
	}
	template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
	char __sr[1<<21],__z[20];int __C=-1,__zz=0;
	inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
	inline void print(register int x)
	{
		if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
		while(__z[++__zz]=x%10+48,x/=10);
		while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
	}
	void file()
	{
		#ifdef NTFOrz
		freopen("a.in","r",stdin);
		#endif
	}
	inline void chktime()
	{
		#ifdef NTFOrz
		cout<<(clock()-t)/1000.0<<'\n';
		#endif
	}
	#ifdef mod
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
	ll inv(ll x){return ksm(x,mod-2);}
	#else
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
	#endif
//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

int n;
char a[sz];
int p1[sz],p2[sz],c1,c2;

string s[sz],dp[sz];
int link[sz],to[sz];

int book[sz],T;

int main()
{
	file();
	read(n);n*=2;
	cin>>(a+1);
	rep(i,1,n) if (a[i]=='a') p1[++c1]=i; else p2[++c2]=i;
	rep(i,1,n>>1) link[min(p1[i],p2[i])]=max(p1[i],p2[i]);
	rep(i,1,n) if (link[i])
	{
		if (a[i]=='a') { s[i]="ab",to[i]=link[i]; continue; }
		++T;book[i]=book[link[i]]=T;int R=link[i];
		rep(j,i,R) if (link[j]) book[j]=book[R=link[j]]=T;
		rep(j,i,R) if (book[j]==T) s[i]+=a[j];
		to[i]=R;
	}
	drep(i,n,1)
	{
		dp[i]=dp[i+1];
		if (link[i]) chkmax(dp[i],s[i]+dp[to[i]+1]);
	}
	cout<<dp[1];
	return 0;
}