Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.

When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.

Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.

Now, he goes through the following process:

* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.



Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).

Constraints

* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)

Input

Input is given from Standard Input in the following format:


N M


Output

Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.

Examples

Input

1 1


Output

3800


Input

10 2


Output

18400


Input

100 5


Output

608000
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		long N = sc.nextLong();
		long M = sc.nextLong();
		long perone = 1900*M+100*(N-M);
		long pgaku = pow(2, M);
		System.out.println(perone*pgaku);
	}
	
	static double kakuritsu(int i, double p) {
		double ret = p;
		for(int j=1;j<i; j++)  {
			ret *= (1-p);
		}
		return ret;
	}
	
	static int binarySearch(int[] A, int x) {
		int l = 0;
		int r = A.length;
		while(l<r) {
			int mid = (l+r)/2;
			if(A[mid] == x) return mid;
			else if(x<A[mid]) r = mid;
			else l = mid+1;
		}
		return -1;
	}
	
	static long upper_bound(int[] A, int b) {
		// bよりおおきい値がはじめて出てくるindexを返す
		return lower_bound(A, b+1);
	}

	static long lower_bound(int[] A, int b) {
		// b以上の値がはじめて出てくるindexを返す
		int lb=0;
		int ub=A.length;
		while(ub>lb) {
			int mid = (lb+ub)/2;
			if(A[mid]>=b) ub = mid;
			else lb = mid+1;
		}
		return lb;
	}
	
	// union find lib
	// usage:
	// 最初にinitを呼ぶ
	// root: 直接は呼ばないで
	// unite: まとめる
	// same: グループ判定
	static void init(int par[], int N) {
		for(int i=0; i<N; i++) {
			par[i] = i;
		}
	}
	
	static int root(int x, int [] par) {
		if(par[x] == x) {
			return x;
		} else {
			return (par[x] = root(par[x], par));
		}
	}
	
	static boolean same(int x, int y, int[] par) {
		return root(x, par) == root(y, par);
	}
	
	static void unite(int x, int y, int[] par) {
		x = root(x, par);
		y = root(y, par);
		if(x == y) return;
		par[x] = y;
	}
	
	// end union find lib
	
	// number lib
	
	static int max(int...ls) {
		int ans = Integer.MIN_VALUE;
		for(int i=0; i<ls.length; i++) {
			ans = Math.max(ans, ls[i]);
		}
		return ans;
	}
	
	static int min(int...ls) {
		int ans = Integer.MAX_VALUE;
		for(int i=0; i<ls.length; i++) {
			ans = Math.min(ans, ls[i]);
		}
		return ans;
	}
	
	static long lcm(long a, long b) {
		return a*(b/gcd(a, b));
	}
	
	static long gcd(long a, long b) {
		long ta = Math.max(a, b);
		long tb = Math.min(a, b);
		long tmp;
		while((tmp = ta%tb) != 0) {
			ta = tb;
			tb = tmp;
		}
		return tb;
	}
	
	static long modcomb(long n, long k, long mod) {
		if(k==1) {
			return n;
		}
		
		long ans = 1;
		for(long i=n; i>=n-k+1; i--) {
			ans = (ans * i)%mod;
		}
		for(long i=k; 0<i; i--) {
			ans = (ans * modpow(i, mod-2, mod)) % mod;
		}
		return ans;
	}
	
	static long modpow(long a, long b, long mod) {
		if(b==0) return 1;
		if(b%2==0) {
			long d = modpow(a, b/2, mod);
			return (d*d)%mod;
		} else {
			return (a*modpow(a, b-1, mod))%mod;
		}
	}
	
	static long pow(long a, long b) {
		if(b==0) return 1;
		if(b==1) return a;
		if(b%2==0) {
			long d = pow(a, b/2);
			return d*d;
		} else {
			return a*pow(a, b-1);
		}
	}
	
	static int disit(long a, long d) {
		int count = 0;
		while(a!=0) {
			a = a/d;
			count++;
		}
		return count;
	}
	
	// end number lib
}