Sequence with Digits

Let’s define the following recurrence : an+1=an+minDigit(an)⋅maxDigit(an).an+1=an+minDigit(an)⋅maxDigit(an).Here minDigit(x)minDigit(x) and maxDigit(x)maxDigit(x) are the minimal and maximal digits in the decimal representation of x without leading zeroes.
For examples refer to notes.Your task is calculate aK for given a1 and K.
Input
The first line contains one integer t(1≤t≤1000) — the number of independent test cases.
Each test case consists of a single line containing two integers a1 and K (1≤a1≤10¹⁸, 1≤K≤10¹⁶) separated by a space.
Output
For each test case print one integer aKaK on a separate line.
Example
input
8
1 4
487 1
487 2
487 3
487 4
487 5
487 6
487 7
ootput
42
487
519
528
544
564
588
628
Note
a1=487
a2=a1+minDigit(a1)⋅maxDigit(a1)=487+min(4,8,7)⋅max(4,8,7)=487+4⋅8=519
a3=a2+minDigit(a2)⋅maxDigit(a2)=519+min(5,1,9)⋅max(5,1,9)=519+1⋅9=528
a4=a3+minDigit(a3)⋅maxDigit(a3)=528+min(5,2,8)⋅max(5,2,8)=528+2⋅8=544
a5=a4+minDigit(a4)⋅maxDigit(a4)=544+min(5,4,4)⋅max(5,4,4)=544+4⋅5=564
a6=a5+minDigit(a5)⋅maxDigit(a5)=564+min(5,6,4)⋅max(5,6,4)=564+4⋅6=588
a7=a6+minDigit(a6)⋅maxDigit(a6)=588+min(5,8,8)⋅max(5,8,8)=588+5⋅8=628

#include<stdio.h>
int fmax(int n)
{
 int res=n%10;
 while(n)
 {
  if(res<n%10)
  {
   res=n%10;
  }
  n=n/10;
 }
 return res;
}
int fmin(int n)
{
 int res=n%10;
 while(n)
 {
  if(res>n%10)
  {
   res=n%10;
  }
  n=n/10;
 }
 return res;
}
int fmn(int n)
{
 return n+fmax(n)*fmin(n);
}
int main()
{
 int t=0;
 scanf("%d",&t);
 while(t--)
 {
  int a1,k=0;
  scanf("%d%d",&a1,&k);
  for(int i=1;i<k;i++)
  {
   a1=fmn(a1);
  }
  printf("%d\n",a1);
 }
}