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)

	//k_k_dubey_
#include<bits/stdc++.h>
#define mod 1000000007
#define ll long long
#define inf 999999999999
#define mx 200005
#define gcd __gcd
#define pb push_back
#define mk make_pair
#define bs binary_search
#define ub upper_bound
#define lb lower_bound
#define endl "\n"
#define vl vector<ll>
#define vs vector<string>
#define non __builtin_popcount
#define ml map<ll, ll> 
#define ms map<string , ll> 
using namespace std;

bool isp[mx];

void sv(){

	ll i,j;
	memset(isp, true, sizeof(isp));
	
	isp[1]= false;
	for(i=2;i*i<mx;i++){
		if(isp[i]){
			for(j=i*i;j<mx;j+=i)
			isp[j]= false;
		}
	}
}

ll powr(ll a,ll b)
{
	ll res=1;
	while(b)
	{
		if(b&1)res=(res*a)%mod;
		a=(a*a)%mod;
		b>>=1;
	}
	return res;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
	//sv();
	
	ll n,k,i,j;
	cin>>n>>k;
	ll ans;
	if(k==1)
	ans= 0;
	else{
	    ans= n-k;
	}
	cout<<ans;
    return 0;
}