CF 339A Helpful Maths

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我是靠谱客的博主眼睛大冰棍，最近开发中收集的这篇文章主要介绍CF 339A Helpful Maths，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

题意：

输入数字和符号‘+’，要求排序后从小到大重新输入；

长度是len，除掉符号数，把数字强制转换后存入另外一个数组，输出的时候，

小技巧：先输入一个数字，再输入加号和数字；

这样可以符合例题的第三个样例；

Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.

The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.

You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.

Input

The first line contains a non-empty string s — the sum Xenia needs to count. String s contains no spaces. It only contains digits and characters "+". Besides, string s is a correct sum of numbers 1, 2 and 3. String s is at most 100 characters long.

Output

Print the new sum that Xenia can count.

Examples

Input

3+2+1

Output

1+2+3

Input

1+1+3+1+3

Output

1+1+1+3+3

Input

Output

AC代码：

#include <bits/stdc++.h>
using namespace std ;
int main()
{
char a[5000];
int b[5000];
cin>>a;
int len
= strlen(a);
int sum ;
sum=0;
for(int i = 0 ; i < len ; i++)
{
if(a[i]=='+')
{
sum++;
}
}
int j ;
j=0;
for(int i = 0 ; i < len ; i++)
{
if(a[i]!='+')
{
b[j]=a[i]-'0';
j++;
}
}
sort(b,b+j);
printf("%d",b[0]);
for(int i = 1 ; i < j ;i++)
{
printf("+%d",b[i]);
}
return 0;
}