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>
int main() {
  int dp[21][5][2][11] = {0};
  dp[2][2][0][0] = 1;
  dp[2][3][0][0] = 2;
  dp[2][4][0][0] = 3;
  for (int i = 3; i < 21; i++) {
    for (int j = 1; j <= 4; j++) {
      for (int k = 1; k < j; k++) {
        for (int o = 0; o <= 10; o++) {
          dp[i][j][0][o] += dp[i - 1][k][1][o];
          dp[i][j][0][o] += dp[i - 1][k][0][o];
        }
      }
      for (int k = 4; k > j; k--) {
        for (int o = 0; o <= 10; o++) {
          dp[i][j][1][o + 1] += dp[i - 1][k][0][o];
          dp[i][j][1][o] += dp[i - 1][k][1][o];
        }
      }
    }
  }
  int n, t;
  while (scanf("%d%d", &n, &t) != EOF) {
    int ans = 0;
    for (int i = 1; i <= 4; i++) {
      ans += dp[n][i][1][t];
    }
    printf("%d\n", ans);
  }
}
