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).
n,t=map(int,input().split())
t=2*t-1
dp=[[[[0,0] for h in range(5)] for j in range(t+2)] for i in range(n+1)]
for h in range(1,5):
    dp[2][0][h][0]=4-h
    dp[2][0][h][1] = h-1

for i in range(2,n):
    for j in range(t+1):
        for h in range(1,5):
            for ne in range(1,5):
                for k in range(2):
                    if (h!=ne):
                        if (k%2):
                            if (ne<h):
                                if (j%2==0):

                                    dp[i+1][j+1][ne][1-k]+=dp[i][j][h][k]
                            else:
                                dp[i+1][j][ne][k]+=dp[i][j][h][k]
                        else:
                            if (ne>h):
                                if (j%2==1):
                                    dp[i+1][j+1][ne][1-k]+=dp[i][j][h][k]
                            else:
                                dp[i+1][j][ne][k]+=dp[i][j][h][k]





ans=0
for h in range(1,4):
    ans+=dp[n][t][h][0]
print(ans)





