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).
#include <bits/stdc++.h>
using namespace std;
int n, t, d[30][15][15][5][5], i, j, l, a, b, s, S;
int main() {
  cin >> n >> t;
  for (i = 1; i < 5; i++)
    for (j = 1; j < 5; j++)
      if (i != j) d[2][0][0][i][j] = 1;
  for (i = 3; i <= n; i++)
    for (j = 0; j <= t; j++)
      for (l = 0; l <= t; l++)
        for (a = 1; a <= 4; a++)
          for (b = 1; b <= 4; b++) {
            S = 0;
            for (s = 1; s <= 4; s++) {
              if (s != a) {
                if (a > s && a > b)
                  S += d[i - 1][j - 1][l][s][a];
                else if (a < s && a < b)
                  S += d[i - 1][j][l - 1][s][a];
                else
                  S += d[i - 1][j][l][s][a];
              }
            }
            if (a != b) d[i][j][l][a][b] += S;
          };
  S = 0;
  for (i = 1; i <= 4; i++)
    for (j = 1; j <= 4; j++) S += d[n][t][t - 1][i][j];
  cout << S;
  return 0;
}
