prison

There are an infinite number of prisoners. Initially, the prisoners are numbered 0, 1, 2, ....

Do the following N times:

* Release the 0th prisoner and execute the k, 2k, 3k, ... th prisoners.
* Then renumber the remaining prisoners. At this time, the prisoners with the smallest original numbers are numbered 0, 1, 2, ... in order.



Find the number that was originally assigned to the prisoner released in the Nth operation.

Constraints

* 1 ≤ N ≤ 10 ^ 5
* 2 ≤ k ≤ 10 ^ 5
* The answer is less than 10 ^ {18}.



Input Format

Input is given from standard input in the following format.


N k


Output Format

Print the answer in one line.

Sample Input 1


4 2


Sample Output 1


7


Sample Input 2


13


Sample Output 2


0


Sample Input 3


100000 100000


Sample Output 3


99999






Example

Input

4 2


Output

7
#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
const double EPS = 1e-9;
typedef pair<int, int> P;
typedef unsigned int ui;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef long double ld;
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
int main() {
	int n; ll k; cin >> n >> k;
	ll out = 0;
	rep(i, n - 1) {
		ll r = out / (k - 1); ll u = out % (k - 1);
		out = r * k + u + 1;
	}
	cout << out << endl;
	return 0;
}
