codeforces B. Minimum Ternary String

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 $j<i$

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

qwq;

这个题就是将1插进字符串里面；2前面必定都是1

代码如下、

#include 
typedef long long ll;
using namespace std;
int main()
{
    string s;
    cin >> s;
    ll one = 0;
    for(int i = 0; i < s.length(); i ++)
    {
        if(s[i] == '1')
            one++;
    }
    for(int i = 0; i < s.length(); i ++)
    {
        if(s[i] == '0')
            cout << 0;
        if(s[i] == '2')
        {
            while(one)
            {
                one--;
                cout << 1;
            }
            cout << 2;
        }
    }
    while(one)
    {
        one--;
        cout << 1;
    }
    cout << endl;
    return 0;
}