Bob likes to draw camels: with a single hump, two humps, three humps, etc. He draws a camel by connecting points on a coordinate plane. Now he's drawing camels with t humps, representing them as polylines in the plane. Each polyline consists of n vertices with coordinates (x1, y1), (x2, y2), ..., (xn, yn). The first vertex has a coordinate x1 = 1, the second — x2 = 2, etc. Coordinates yi might be any, but should satisfy the following conditions:

  * there should be t humps precisely, i.e. such indexes j (2 ≤ j ≤ n - 1), so that yj - 1 < yj > yj + 1, 
  * there should be precisely t - 1 such indexes j (2 ≤ j ≤ n - 1), so that yj - 1 > yj < yj + 1, 
  * no segment of a polyline should be parallel to the Ox-axis, 
  * all yi are integers between 1 and 4. 



For a series of his drawings of camels with t humps Bob wants to buy a notebook, but he doesn't know how many pages he will need. Output the amount of different polylines that can be drawn to represent camels with t humps for a given number n.

Input

The first line contains a pair of integers n and t (3 ≤ n ≤ 20, 1 ≤ t ≤ 10).

Output

Output the required amount of camels with t humps.

Examples

Input

6 1


Output

6


Input

4 2


Output

0

Note

In the first sample test sequences of y-coordinates for six camels are: 123421, 123431, 123432, 124321, 134321 и 234321 (each digit corresponds to one value of yi).
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF14E extends PrintWriter {
	CF14E() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF14E o = new CF14E(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int t = sc.nextInt();
		long[][][][][] dp = new long[n][t + 1][t][4][4];
		for (int y0 = 0; y0 < 4; y0++)
			for (int y1 = 0; y1 < 4; y1++)
				if (y0 != y1)
					dp[1][0][0][y0][y1] = 1;
		for (int i = 1; i < n - 1; i++)
			for (int h = 0; h <= t; h++)
				for (int d = 0; d < t; d++)
					for (int y0 = 0; y0 < 4; y0++)
						for (int y1 = 0; y1 < 4; y1++) {
							long x = dp[i][h][d][y0][y1];
							if (x == 0)
								continue;
							for (int y2 = 0; y2 < 4; y2++) {
								if (y2 == y1)
									continue;
								if (y0 < y1 && y1 < y2 || y0 > y1 && y1 > y2) {
									dp[i + 1][h][d][y1][y2] += x;
								} else if (y0 < y1 && y1 > y2) {
									if (h + 1 <= t)
										dp[i + 1][h + 1][d][y1][y2] += x;
								} else if (y0 > y1 && y1 < y2) {
									if (d + 1 < t)
										dp[i + 1][h][d + 1][y1][y2] += x;
								}
							}
						}
		long ans = 0;
		for (int y0 = 0; y0 < 4; y0++)
			for (int y1 = 0; y1 < 4; y1++)
				ans += dp[n - 1][t][t - 1][y0][y1];
		println(ans);
	}
}
