Anton has a positive integer n, however, it quite looks like a mess, so he wants to make it beautiful after k swaps of digits.
Let the decimal representation of n as (x1x2… xm)10 satisfying that 1 ≤ x1 ≤ 9, 0 ≤ xi ≤ 9 (2 ≤ i ≤ m), which means . In each swap, Anton can select two digits xi and xj (1 ≤ i ≤ j ≤ m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after k swaps?
Input
The first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1 ≤ T ≤ 100, 1 ≤ n, k ≤ 109.
Output
For each test case, print in one line the minimum integer and the maximum integer which are separated by one space.
Example
Input
5
12 1
213 2
998244353 1
998244353 2
998244353 3
Output
12 21
123 321
298944353 998544323
238944359 998544332
233944859 998544332
Tips
给一个数,求你k次交换能得到的最大数和最小数,用深搜求全排列加减枝
#include<bits/stdc++.h>
#define read() freopen("input.txt","r",stdin);
#define write() freopen("output.txt","w",stdout);
void FILE_IO(){
#ifndef ONLINE_JUDGE
read();write();
#endif
}
using namespace std;
string mini,maxi,line;
void bt(int pos,int taken){
if(pos>=line.size()){
cout<<line<<'\n';
maxi=max(maxi,line);
if(line[0]!='0')
mini=min(mini,line);
return;
}
bt(pos+1,taken);
if(taken-1>=0){
for( int i=pos+1; i<line.size(); i++ ){
swap(line[i],line[pos]);
bt(pos+1,taken-1);
swap(line[i],line[pos]);
}
}
}
int main() {
FILE_IO();
read();
int t,k;
cin>>t;
while(t--){
cin>>line>>k;
mini=line;
maxi=line;
bt(0,k);
cout<<mini<<' '<<maxi<<endl;
}
}