Min Number

Now you are given one non-negative integer n in 10-base notation, it will only contain digits ('0'-'9'). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].

For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.

Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?

Please note that in this problem, leading zero is not allowed!

Input

The first line of the input contains an integer T (T≤100), indicating the number of test cases.

Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.

Output

For each test case, output the minimum number we can get after no more than M operations.

Sample Input

3

9012 0

9012 1

9012 2

Sample Output

9012

1092

1029

题意：

输入一个数，和n操作步数

交换n次操作步数，使这个数尽可能的变小

注意：第一个数字不能为0

#include<stdio.h>#include<string.h>int main(){int i,j,t,n,mini,x,len;char str[1004],min;while(scanf("%d",&t)!=EOF){  getchar();  while(t--)  {   scanf("%s%d",str,&n);   len=strlen(str);            for(i=0;i<len;i++)    {      if(n==0) break;      min=str[i];x=mini=i;      for(j=i+1;j<len;j++)                      {         if(i==0)//考虑第一个不为0           {            if(min>str[j]&&str[j]!='0') {mini=j;min=str[j];}                 }                                         else if(min>str[j]){mini=j;min=str[j];}       }          int k;                    if(mini!=x) {k=str[mini];str[mini]=str[x];str[x]=k;n--;}    }              printf("%s\n",str);  }                          }    return 0;}