B. Minimum Ternary String

链接：https://codeforces.com/problemset/problem/1009/B

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

Copy

100210

output

Copy

001120

input

Copy

11222121

output

Copy

11112222

input

Copy

20

output

Copy

20

代码：

#include <bits/stdc++.h>
using namespace std;
long long t,n,m,k,y,s=0,sum=0,min1=0,max1=0;
char x[100005];
long long s1=0,s2=0,s0=0;
long long p[100005];
int main()
{
    cin>>x;
    n=strlen(x);
    int j=0;
    for(int i=0;i<n;i++)
    {
    	if(x[i]=='2')
    	s2++;
    	if(x[i]=='1')
    	s1++;
    	if(x[i]=='0')
    	s0++;
    	if(x[i]=='2')
    	{
    		j++;
    		p[j]=s0;
    		s0=0;
    	}
	}
	if(j>=1)
	{
		for(int i=1;i<=p[1];i++)
	    cout<<0;
	}
    else
    {
    	for(int i=1;i<=s0;i++)
	    cout<<0;
    }
    for(int i=1;i<=s1;i++)
    cout<<1;
    if(s2>=1)
    {
    	cout<<2;
    	s2--;
    }
    for(int i=2;i<=j;i++)
    {
    	for(int o=1;o<=p[i];o++)
    	{
    		cout<<0;
    	}
    	if(s2>=1)
	    {
	    	cout<<2;
	    	s2--;
	    }
    }
    if(j>=1)
    {
	    for(int i=1;i<=s0;i++)
	    cout<<0;
    }
    
}