Problem: 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 '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication: 

<image>

<image>

<image>

<image>

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

<image>. 

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

<image>.

For the given logical expression <image> 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.

Input

The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.

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

Output

Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.

Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.

The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.

The expression should be correct. More formally, a correct expression is determined as follows:

  * Expressions "0", "1" (without the quotes) are correct. 
  * If v1, v2 are correct, then v1->v2 is a correct expression. 
  * If v is a correct expression, then (v) is a correct expression. 



The total number of characters in the resulting expression mustn't exceed 106.

If there are multiple possible answers, you are allowed to print any of them.

Examples

Input

4
0 1 1 0


Output

YES
(((0)-&gt;1)-&gt;(1-&gt;0))


Input

2
1 1


Output

NO


Input

1
0


Output

YES
0

Solution (in JAVA):

Solution:
/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.Arrays;

/* Name of the class has to be "Main" only if the class is public. */
public class ValentineProject {

    private void solve() throws IOException {
    Deque<Character> cnt = new ArrayDeque<Character>();

    int a[] = new int[100005];

	int n = nextInt();
	for(int i=1;i<=n;++i)
		a[i] = nextInt();
    
    if( a[n] == 1){
		writer.println("NO");
		return ;
	}
	if( n == 1){
		writer.println("YES\n0");
		return ;
	}
	if( a[n-1] == 0){
		boolean zero = false;
		for(int i=1;i<n-1;++i)
			if( a[i] == 0){
				zero = true;
				break;
			}
		if( zero == false){
			writer.println("NO");
			return ;
		}
	}
	
	if( a[n-1] == 0)
		cnt.addFirst('0');
	else
		cnt.addFirst('1');
		
	for(int i = n -2 ; i >=1; i--){
		cnt.addFirst( '>');
		cnt.addFirst( '-');
		if( a[i] == 0)
			cnt.addFirst('0');
		else
			cnt.addFirst('1');
		cnt.addFirst( '(');
		cnt.addLast(')');
	}

	cnt.addLast('-');
	cnt.addLast('>');
	cnt.addLast('0');
	System.out.println("YES");
	while( cnt.size() > 0 ){
		System.out.print(cnt.getFirst());
		cnt.remove();
	}
	
    


    }
    
    
    public static void main(String[] args) {
        // TODO code application logic here
        new ValentineProject().run();
    }
    
    BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter writer;
    
    public void run() {
        try {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            writer = new PrintWriter(System.out);
            solve();
            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }
    
    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }
    
}