Problem: Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, …, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.

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It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 10^5) — the amount of numbers on the paper.

The second line of the input contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 9), where a_i is the i-th element of a.

The third line of the input contains the string s (1 ≤ |s| ≤ 3) — symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.

Output

Print n numbers separated by n - 1 symbols — a mathematical expression with the greatest result. If there are multiple equally valid results — output any one of them.

Examples

Input


3
2 2 0
+-*


Output


2*2-0


Input


4
2 1 1 2
+*


Output


2+1+1+2

Note

The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define clr(x) memset(x,0,sizeof(x))
#define For(i,a,b) for (int i=(a);i<=(b);i++)
#define Fod(i,b,a) for (int i=(b);i>=(a);i--)
#define fi first
#define se second
#define kill _z_kill
#define y0 _z_y0
#define y1 _z_y1
#define x0 _z_x0
#define x1 _z_x1
#define pb push_back
#define mp(x,y) make_pair(x,y)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned uint;
typedef long double LD;
typedef vector <int> vi;
typedef pair <int,int> pii;
void enable_comma(){}
string tostring(char c){
	string s="";
	s+=c;
	return s;
}
string tostring(string s){
	return "\""+s+"\"";
}
string tostring(const char *c){
	return tostring((string)c);
}
string tostring(long long x){
	if (x<0)
		return "-"+tostring(-x);
	if (x>9)
		return tostring(x/10)+tostring(char('0'+x%10));
	else
		return tostring(char('0'+x));
}
string tostring(int x){
	return tostring((long long)x);
}
string tostring(unsigned long long x){
	if (x>9)
		return tostring((long long)(x/10))+tostring(char('0'+x%10));
	else
		return tostring(char('0'+x));
}
string tostring(unsigned x){
	return tostring((long long)x);
}
string tostring(double x){
	static char res[114];
	sprintf(res,"%lf",x);
	string s=tostring(res);
	return s.substr(1,(int)s.size()-2);
}
string tostring(long double x){
	return tostring((double)x);
}
template <class A,class B> string tostring(pair <A,B> p){
	return "("+tostring(p.fi)+","+tostring(p.se)+")";
}
template <class T> string tostring(T v){
	string res="";
	for (auto p : v)
		res+=(res.size()?",":"{")+tostring(p);
	return res.size()?res+"}":"{}";
}
template <class A> string tostring(A* a,int L,int R){
	return tostring(vector <A>(a+L,a+R+1));
};
template <class A> string tostring(A a,int L,int R){
	return tostring(a.data(),L,R);
}
string tostrings(){
	return "";
}
template <typename Head,typename... Tail>
string tostrings(Head H,Tail... T){
	return tostring(H)+" "+tostrings(T...);
}
#define User_Time ((double)clock()/CLOCKS_PER_SEC)
#ifdef zzd
	#define outval(x) cerr<<#x" = "<<tostring(x)<<endl
	#define outvals(...) cerr<<"["<<#__VA_ARGS__<<"]: "<<\
								tostrings(__VA_ARGS__)<<endl
	#define outtag(x) cerr<<"--------------"#x"---------------"<<endl
	#define outsign(x) cerr<<"<"#x">"<<endl
	#define outarr(a,L,R) cerr<<#a"["<<(L)<<".."<<(R)<<"] = "<<\
								tostring(a,L,R)<<endl
#else
	#define outval(x) enable_comma()
	#define outvals(...) enable_comma()
	#define outtag(x) enable_comma()
	#define outsign(x) enable_comma()
	#define outarr(a,L,R) enable_comma()
#endif
#ifdef ONLINE_JUDGE
	#ifdef assert
		#undef assert
	#endif
	#define assert(x) (!(x)?\
			cout<<"Assertion failed!"<<endl<<\
			"function: "<<__FUNCTION__<<endl<<\
			"line: "<<__LINE__<<endl<<\
			"expression: "<<#x<<endl,exit(3),0:1)
#endif
LL read(){
	LL x=0,f=0;
	char ch=getchar();
	while (!isdigit(ch))
		f=ch=='-',ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return f?-x:x;
}
template <class T> void ckmax(T &x,const T y){
	if (x<y)
		x=y;
}
template <class T> void ckmin(T &x,const T y){
	if (x>y)
		x=y;
}
//int Pow(int x,LL y){
//	y=(y%(mod-1)+(mod-1))%(mod-1);
//	int ans=1;
//	for (;y;y>>=1,x=(LL)x*x%mod)
//		if (y&1)
//			ans=(LL)ans*x%mod;
//	return ans;
//}
//void Add(int &x,int y){
//	if ((x+=y)>=mod)
//		x-=mod;
//}
//void Del(int &x,int y){
//	if ((x-=y)<0)
//		x+=mod;
//}
//int Add(int x){
//	return x>=mod?x-mod:x;
//}
//int Del(int x){
//	return x<0?x+mod:x;
//}
//int md(LL x){
//	return (x%mod+mod)%mod;
//}
const int N=100005,INF=2e6;
int n;
int a[N];
int p[N],cnt=0;
int dp[100],pre[100];
int vv[N];
void solve(int L,int R){
	assert(L<=R);
	if (L==R)
		return printf("%d",a[L]),void();
	if (L<R&&a[L]==1){
		printf("1+");
		solve(L+1,R);
		return;
	}
	if (L<R&&a[R]==1){
		solve(L,R-1);
		printf("+1");
		return;
	}
	int v=1;
	For(i,L,R)
		v=min(v*a[i],INF);
	if (v>=INF){
		For(i,L,R-1)
			printf("%d*",a[i]);
		printf("%d",a[R]);
		return;
	}
	cnt=0;
	For(i,L,R)
		if (a[i]>1)
			p[++cnt]=i;
	p[cnt+1]=R+1;
	clr(dp);
	dp[0]=0;
	For(i,0,cnt-1){
		int v=1;
		For(j,i+1,cnt){
			v*=a[p[j]];
			int val=dp[i]+v+p[j+1]-p[j]-1;
			if (val>dp[j]){
				dp[j]=val;
				pre[j]=i;
			}
		}
	}
	for (int i=cnt;i;i=pre[i]){
		For(j,p[pre[i]+1],p[i]-1)
			vv[j]=1;
	}
	For(i,L,R-1)
		printf("%d%c",a[i],vv[i]==0?'+':'*');
	printf("%d",a[R]);
}
void so1(int L,int R){
	if (L==R)
		return printf("%d",a[L]),void();
	if (L<R&&a[L]==0){
		printf("0+");
		so1(L+1,R);
		return;
	}
	if (L<R&&a[R]==0){
		so1(L,R-1);
		printf("+0");
		return;
	}
	vi p;
	For(i,L,R)
		if (!a[i])
			p.pb(i);
	if (p.empty())
		solve(L,R);
	else {
		solve(L,p[0]-1);
		printf("+0+");
		For(i,0,(int)p.size()-2){
			if (p[i]+1<=p[i+1]-1){
				solve(p[i]+1,p[i+1]-1);
				printf("+0+");
			}
			else {
				printf("0+");
			}
		}
		solve(p.back()+1,R);
	}
}
int main(){
	n=read();
	For(i,1,n)
		a[i]=read();
	string s,t;
	cin>>s;
	int flag=0;
	for (auto i : s)
		if (i!='-')
			t+=i;
		else
			flag=1;
	if (s.size()==1){
		For(i,1,n)
			printf("%d%c",a[i],i==n?'\n':s[0]);
		return 0;
	}
	if (t.size()==1&&(t=="+"||!flag||(t=="*"&&flag&&a[1]==0))){
		For(i,1,n)
			printf("%d%c",a[i],i==n?'\n':t[0]);
		return 0;
	}
	if (t.size()==1&&flag){
		assert(t=="*");
		int L=1;
		while (L<=n&&a[L]>0){
			if (L>1)
				putchar('*');
			printf("%d",a[L++]);
		}
		if (L>n)
			return 0;
		For(i,L,n)
			printf("%c%d",i==L?'-':'*',a[i]);
		return 0;
	}
	so1(1,n);
	puts("");
	return 0;
}