A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.

We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2

-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3

.

Now, you are given RBS s

of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b

. No brackets can be left uncolored.

Among all possible variants you should choose one that minimizes maximum of r

's and b

's nesting depth. If there are multiple solutions you can print any of them.

Input

The first line contains an even integer n

(2≤n≤2⋅105) — the length of RBS s

.

The second line contains regular bracket sequence s

(|s|=n, si∈{"(", ")"}

).

Output

Print single string t

of length n consisting of "0"-s and "1"-s. If ti is equal to 0 then character si belongs to RBS r, otherwise si belongs to b

.

Examples

Input

Copy

2
()

Output

Copy

11

Input

Copy

4
(())

Output

Copy

0101

Input

Copy

10
((()())())

Output

Copy

0110001111

Note

In the first example one of optimal solutions is s=

"()". r is empty and b= "()". The answer is max(0,1)=1

.

In the second example it's optimal to make s=

"(())". r=b= "()" and the answer is 1

.

In the third example we can make s=

"((()())())". r= "()()" and b= "(()())" and the answer is 2

.


直接模拟，所有左括号按10101010。。。排列，右括号与左边离自己最近且未匹配的左括号匹配

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<set>
#define ll long long
#define mod 998244353
using namespace std;
char a[205000];
int b[200500];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",a+1);
        int m=0;
        for(int i=1;i<=n;i++)
        {
            if(a[i]=='(')
            {
                m=(m+1)%2;
                b[i]=m;
            }
            else
            {
                b[i]=m;
                m=(m+1)%2;
            }
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d",b[i]);
        }
        printf("\n");
    }
}