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 read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
inline void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
  return;
}
int n, f[20 + 5][10 + 5][5 + 5][2];
int main() {
  int n = read(), T = read();
  for (int i = 1; i <= 4; i++) f[0][0][i][0] = 1;
  for (int x = 1; x < n; ++x)
    for (int t = 0; t <= T; ++t)
      for (int i = 1; i <= 4; ++i)
        for (int j = 1; j <= 4; ++j) {
          if (j < i)
            f[x][t][i][1] += f[x - 1][t][j][1] + f[x - 1][t][j][0];
          else if (j > i && t)
            f[x][t][i][0] += f[x - 1][t][j][0] + f[x - 1][t - 1][j][1];
        }
  int sum = 0;
  for (int i = 1; i <= 4; ++i) sum += f[n - 1][T][i][0];
  printf("%d\n", sum);
  return 0;
}
