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 dp[22][11][11][5][2];
int main() {
  int n, t;
  cin >> n >> t;
  if (n == 1) {
    cout << 0 << endl;
    return 0;
  }
  int i, j, k, l, h, x, y;
  dp[2][0][0][1][0] = 3;
  dp[2][0][0][1][1] = 0;
  dp[2][0][0][2][0] = 2;
  dp[2][0][0][2][1] = 1;
  dp[2][0][0][3][0] = 1;
  dp[2][0][0][3][1] = 2;
  dp[2][0][0][4][0] = 0;
  dp[2][0][0][4][1] = 3;
  for (i = 3; i <= n; i++)
    for (j = t; j >= 0; j--)
      for (k = t - 1; k >= 0; k--) {
        for (x = 1; x <= 4; x++) {
          for (y = 1; y < x; y++) {
            if (k > 0) dp[i][j][k][x][1] += dp[i - 1][j][k - 1][y][0];
            dp[i][j][k][x][1] += dp[i - 1][j][k][y][1];
          }
          for (y = x + 1; y <= 4; y++) {
            if (j > 0) dp[i][j][k][x][0] += dp[i - 1][j - 1][k][y][1];
            dp[i][j][k][x][0] += dp[i - 1][j][k][y][0];
          }
        }
      }
  int rinka = 0;
  for (i = 1; i <= 4; i++)
    for (j = 0; j <= 1; j++) rinka += dp[n][t][t - 1][i][j];
  cout << rinka << endl;
  return 0;
}
