解題思路：先找到第一個2出現之前0的個數並標記第一個二所在位置（0和2不能交換），找到字串中所有1的個數（1可以和0,2交換），把找到的0和1按先後順序輸出，此時就剩下2和其後面的0了，再從第一個二的位置把2和0按順序輸出，此時就是所求的最小字串（2越靠後越好）

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

Input

The first line of the input contains the string ss consisting only of characters '0', '1' and '2', its length is between 11 and 105105 (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

程式碼：

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
using namespace std;
#define maxn 100000
int main()
{
    char s[maxn+10],t;
    long long int i,j,l,m,n,flag;
    while(scanf(" %s",s)!=EOF)
    {
        l=strlen(s);
        m=0,n=0;
        flag=1;
        for(i=0;i<l;i++)
        {
            if(flag==1&&s[i]=='0')
                m++;
            else if(s[i]=='1')
                n++;
            else if(flag==1&&s[i]=='2')
            {
                flag=0;
                j=i;
            }
        }
        for(i=0;i<m;i++)
            printf("0");
        for(i=0;i<n;i++)
            printf("1");
        for(i=j;i<l;i++)
            if(s[i]!='1')
                printf("%c",s[i]);
        printf("\n");
    }
    return 0;
}