Minimum Ternary String （贪心）

You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').

You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).

For example, for string "010210" we can perform the following moves:

• "010210" →

• "100210";
• "010210" →
• "001210";
• "010210" →
• "010120";
• "010210" →

• "010201".

Note than you cannot swap "02" →

"20" and vice versa. You cannot perform any other operations with the given string excluding described above.

You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).

String a

is lexicographically less than string b (if strings a and b have the same length) if there exists some position i (1≤i≤|a|, where |s| is the length of the string s) such that for every j<i holds aj=bj, and ai<bi

.

Input

The first line of the input contains the string s

consisting only of characters '0', '1' and '2', its length is between 1 and 105

(inclusive).

Output

Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).

Examples

Input

100210

Output

001120

Input

11222121

Output

11112222

Input

20

Output

20

题意：将一个只包含012的字符串通过相互交换位置变成字典序最小的字符串，相邻位置下，0和1,1和2可以交换，0和2不能交换

思路：只有0和2不能交换，所以对1来说没有影响，先把所有1的个数记录下来，从前往后

遍历，遇到0输出，遇到1跳过，遇到第一个2，输出所有的1，在输出2，之后的位置遇到0输出，遇到1跳过，遇到2输出

代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main(){
	string str;
	cin>>str;
	int ans1=0,ans2=0,ans0=0;
	for(int i=0;i<str.length();i++){
		ans1 += str[i]=='1';
		ans0 += str[i]=='0';
		ans2 += str[i]=='2';
	}
	if(ans0==0||ans2==0){
		sort(str.begin(),str.end());
		cout<<str<<endl;
		return 0;
	} 
	for(int i=0;i<str.length();i++){
		if(str[i]=='0') cout<<0;
		else if(str[i]=='2'&&ans1){
			while(ans1){
				cout<<1;
				ans1--;
			}
			cout<<2;
		}
		else{
			if(str[i]=='1') continue;
			if(str[i]=='2') cout<<2;
		}
	}
	return 0;
}