Problem:
Score : 100 points

Problem Statement

Takahashi is distributing N balls to K persons.

If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?

Constraints

1 \leq K \leq N \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the maximum possible difference in the number of balls received.

Sample Input 1

3 2

Sample Output 1

1

The only way to distribute three balls to two persons so that each of them receives at least one ball is to give one ball to one person and give two balls to the other person.

Thus, the maximum possible difference in the number of balls received is 1.

Sample Input 2

3 1

Sample Output 2

0

We have no choice but to give three balls to the only person, in which case the difference in the number of balls received is 0.

Sample Input 3

8 5

Sample Output 3

3

For example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of balls received between the person with the most balls and the person with the fewest balls would be 3, which is the maximum result.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for (int (i) = 0; (i) < (int)(n); i++) 
#define FOR(i, a, b) for(int (i) = a; (i) < (int)b; i++)
#define RREP(i, n) for(int (i)=((int)(n)-1); (i)>=0; i--)
#define RFOR(i, a, b) for(int (i) =((int)(b)-1); (i)>=(int)a; i--)
#define ALL(v) (v).begin(),(v).end()
#define MOD 1000000007
#define FI first
#define SE second
#define MP make_pair
#define PB push_back
#define SZ(x)  (int)x.size()
#define SP(x)  setprecision((int)x)


using namespace std ;
typedef long long ll;
typedef vector<int> vint;
typedef vector<vint> vvint;
typedef vector<string> vstr;
typedef pair<int, int> pii;
const int INF = 1e9;
const ll LINF = 1e18;
const double EPS = 1e-9;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}


int main()
{
  int n, k ;
  cin >> n >>  k ;
  
  if(k==1){
    cout << 0 << endl ;
  }
  else{
    cout << n-k <<endl ;
  }


  return 0 ;
}