1009B 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

题意：给你一串字符只有0，1，2三种字符，如果1，2，或者0，1相邻，可以交换位置，只有0，2不能交换位置。输出经过这种方法，字典序最少的字符串。

题解：贪心  因为只有0和2位置不能交换，所以对于1来说没有影响，先把所有1的个数记录下来。然后从前往后遍历，遇见0就输出0，遇见1就跳过，遇见2把全部1输出，再输出2，然后后面的遇见0就输出0，遇见2输出2，遇见1跳过。

c++：

#include
using namespace std;
int num,flag;
int main()
{
    string s,a;
    cin>>s;
    for(int i=0; i

python:

s=input()
ans=''.join(c for c in s if c!='1')
i=(ans+'2').find('2')
print(ans[:i]+s.count('1')*'1'+ans[i:])