Educational Codeforces Round 47 (Rated for Div. 2) 1009B. Minimum Ternary String

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B. Minimum Ternary String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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" $\to$ "100210";
• "010210" $\to$ "001210";
• "010210" $\to$ "010120";
• "010210" $\to$ "010201".

Note than you cannot swap "02" $\to$ "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 $i$ ($1\le i\le |a|$, where $|s|$ is the length of the string $s$) such that for every $j<i$ holds $a_j=b_j$, and $a_i<b_i$.

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 ${10}^5$ (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

题意:

1可以跟0互换位置，2和1可以互换位置，给你一个串，让你通过交换位置，得到字典序最小的串

解析:

在这道题里面，1可以换到任意的位置，因为1<->2,1<->0，而2和0的相对位置固定，例如一个0跟在一个2后面，则这个0无论怎么换都在这个2后面

所以我们只需要把这个串里面的1全部提取出来，全部集中放在前面的一个位置，使得字典序最小。

#include <cstdio>
#include <bitset>
using namespace std;

const int MAXN = 1e5+1;

char str[MAXN];
char s[MAXN];
int main()
{
    scanf("%s",str);
    int cnt=0;
    int an1=0;
    for(int i=0;str[i];i++)
    {
        if(str[i]=='1')
        {
            an1++;
            continue;
        }
        s[cnt++]=str[i];
    }
    int k=0;
    while(s[k]=='0'&&k<cnt)
        printf("0"),k++;
    for(int i=0;i<an1;i++) printf("1");
    if(k<cnt)printf("%s\n",s+k);
    return 0;
}