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 groB(int a, int b, int c) {
  if (a == b || b == c) return -1;
  if (b > c && b > a) return 1;
  return 0;
}
int main() {
  int n, t;
  cin >> n >> t;
  int dp[22][11][4][4];
  for (int i = 0; i < 22; i++)
    for (int j = 0; j < 11; j++)
      for (int q = 0; q < 4; q++)
        for (int w = 0; w < 4; w++) dp[i][j][q][w] = 0;
  for (int i = 0; i < 4; i++)
    for (int j = i + 1; j < 4; j++) dp[1][0][i][j] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j <= t; j++) {
      for (int f = 0; f < 4; f++)
        for (int s = 0; s < 4; s++) {
          dp[i][j][f][s] = 0;
          for (int tt = 0; tt < 4; tt++) {
            int tmp = groB(tt, f, s);
            if (tmp == -1) continue;
            if (tmp)
              dp[i][j][f][s] += dp[i - 1][j - 1][tt][f];
            else
              dp[i][j][f][s] += dp[i - 1][j][tt][f];
          }
        }
    }
  }
  int res = 0;
  for (int i = 1; i < 4; i++)
    for (int j = 0; j < i; j++) res += dp[n - 1][t][i][j];
  cout << res << endl;
}
