【CodeForces 1251C --- Minimize The Integer】

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You are given a huge integer a consisting of n digits (n is between 1 and 3⋅105, inclusive). It may contain leading zeros.

You can swap two digits on adjacent (neighboring) positions if the swapping digits are of different parity (that is, they have different remainders when divided by 2).

For example, if a=032867235 you can get the following integers in a single operation:

• 302867235 if you swap the first and the second digits;
• 023867235 if you swap the second and the third digits;
• 032876235 if you swap the fifth and the sixth digits;
• 032862735 if you swap the sixth and the seventh digits;
• 032867325 if you swap the seventh and the eighth digits.

Note, that you can’t swap digits on positions 2 and 4 because the positions are not adjacent. Also, you can’t swap digits on positions 3 and 4 because the digits have the same parity.

You can perform any number (possibly, zero) of such operations.

Find the minimum integer you can obtain.

Note that the resulting integer also may contain leading zeros.

Input

The first line contains one integer t (1≤t≤104) — the number of test cases in the input.

The only line of each test case contains the integer a, its length n is between 1 and 3⋅105, inclusive.

It is guaranteed that the sum of all values n does not exceed 3⋅105.

Output

For each test case print line — the minimum integer you can obtain.

Sample Input

3
0709
1337
246432

Sample Output

0079
1337
234642

Note

In the first test case, you can perform the following sequence of operations (the pair of swapped digits is highlighted): 070−−9→0079.

In the second test case, the initial integer is optimal.

In the third test case you can perform the following sequence of operations: 24643−−2→2463−−42→243−−642→234642.

解题思路：

根据题意我们可以理解为相同奇偶性的数字顺序是不能变的，能变得只是不同奇偶性的数字。
所以我们可以通过两个队列来分别按顺序存储奇数和偶数，然后判断大小依次输出就行。

AC代码：

#include <iostream>
#include <string>
#include <queue>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

int main()
{
    SIS;
    string s;
    queue<char> q1,q2;
    int T;
    cin >> T;
    while(T--)
    {
        string ans;
        cin >> s;
        int len=s.size();
        for(int i=0;i<len;i++)
        {
            if(s[i]&1) q1.push(s[i]);
            else q2.push(s[i]);
        }
        while(!q1.empty() || !q2.empty())
        {
            if(q1.empty() || (!q2.empty() && q1.front()>q2.front()))
            {
                ans.push_back(q2.front());
                q2.pop();
            }
            else
            {
                ans.push_back(q1.front());
                q1.pop();
            }
        }
        cout << ans << endl;
    }
    return 0;
}