Educational Codeforces Round 43 (Rated for Div. 2) A. Minimum Binary Number

A. Minimum Binary Number

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".

You are given a correct string s.

You can perform two different operations on this string:

1. swap any pair of adjacent characters (for example, "101"  "110");
2. replace "11" with "1" (for example, "110"  "10").

Let val(s) be such a number that s is its binary representation.

Correct string a is less than some other correct string b iff val(a) < val(b).

Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).

Input

The first line contains integer number n (1 ≤ n ≤ 100) — the length of string s.

The second line contains the string s consisting of characters "0" and "1". It is guaranteed that the string s is correct.

Output

Print one string — the minimum correct string that you can obtain from the given one.

Examples

input

Copy

4
1001

output

Copy

100

input

Copy

1
1

output

Copy

1

Note

In the first example you can obtain the answer by the following sequence of operations: "1001"  "1010"  "1100"  "100".

In the second example you can't obtain smaller answer no matter what operations you use.

题解：
因为不存在前导0，因此第一位一定是1,后面的1一定可以移到前面
与前面的1抵消，最后只剩一个1。
代码：
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=104;
char t[maxn];
int main()
{
	int n;scanf("%d",&n);
	scanf("%s",&t);
	int x1=0,x2=0;
	for(int i=0;i<n;i++)
	{
		if(t[i]=='0')x1++;
		else x2++;
	}
	for(int i=0;i<min(x2,1);i++)printf("1");
	for(int i=0;i<x1;i++)printf("0");
	return 0;
}