Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.

Implication is written by using character '', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:

When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,

.

When there are brackets, we first calculate the expression in brackets. For example,

.

For the given logical expression determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.

The second line contains n numbers a1, a2, ..., an (), which means the values of arguments in the expression in the order they occur.

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define test freopen("test.txt","r",stdin)

#define maxn 2000001

#define mod 10007

#define eps 1e-9

int Num;

char CH[];

const int inf=0x3f3f3f3f;

const ll infll = 0x3f3f3f3f3f3f3f3fLL;

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

inline void P(int x)

{

    Num=;if(!x){putchar('');puts("");return;}

    while(x>)CH[++Num]=x%,x/=;

    while(Num)putchar(CH[Num--]+);

    puts("");

}

//**************************************************************************************

char num[];

int main(){

    int n;

    scanf(" %d",&n);

    for(int i=;i<n;++i)

        scanf(" %c",&num[i]);

    if( n== ){

        if( num[]=='' ) printf("YES\n0\n");

        else printf("NO\n");

        return ;

    }

    if( n== ){

        if( num[]=='' && num[]=='' )

            printf("YES\n1->0\n");

        else printf("NO\n");

        return ;

    }

    if( num[n-]=='' ){

        printf("NO\n");

        return ;

    }

    if( num[n-]=='' ){

        printf("YES\n");

        for(int i=;i<n-;++i)

            printf("%c->",num[i]);

        printf("%c",num[n-]);

        printf("\n");

        return ;

    }

    int cnt =  , from = - , to = n-;

    for(int i=n-;i>= && from==-;--i)

        if( num[i]=='' )

            from = i;

    if( from==- ){

        printf("NO\n");

        return ;

    }

    printf("YES\n");

    for(int i=;i<from;++i)

        printf("%c->",num[i]);

    for(int i=from;i<to;++i)

        printf("(%c->",num[i]);

    printf("%c",num[to]);

    for(int i=from;i<to;++i)

        printf(")");

    printf("->%c\n",num[n-]);

}