Little Praneet loves experimenting with algorithms and has devised a new algorithm. The algorithm is performed on an integer as follows:
- if the rearmost digit is $0$, he will erase it.
- else, he will replace the rearmost digit $d$ with $d-1$.
If a point comes when the integer becomes $0$, the algorithm stops.
You are given an integer $n$. Praneet will perform the algorithm on it $a$ times. You have to print the result after $a$ operations.
-----Input:-----
- The first and only line of input contains two integers $n$ — initial number, and $a$ —the number of operations.
-----Output:-----
- Print one integer — the result of performing the algorithm on $n$ $a$ times.
-----Constraints-----
- $2 \leq n \leq 10^9$
- $1 \leq a \leq 50$
-----Sample Input 1-----
1001 2
-----Sample Input 2-----
5 2
-----Sample Output 1-----
100
-----Sample Output 2-----
3
-----Explanation-----
- In the first example, the transformation is as follows: $1001->1000->100$.
- In the second example, the transformation is as follows: $5->4->3$.
# cook your dish here
n,a=map(int,input().split())
for i in range(a):
 if n!= 0:
  r=n%10
  n=n//10
  if r!=0:
   r-=1
   n=n*10+r
print(int(n))
# cook your dish here
n,a=map(int,input().split())
for i in range(a):
 if n!= 0:
  r=n%10
  n=n//10
  if r!=0:
   r-=1
   n=n*10+r
print(int(n))