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;
#pragma comment(linker, "/STACK:64000000")
const int infi = 1e9 + 7;
const long long infl = 1e18;
const long double eps = 1e-8;
const long double eps1 = 1e-10;
const int mod = (1e9 + 7);
const long long mod1 = 1e9 + 7;
const long long mod2 = 1e9 + 7;
const long long p1 = 31;
const long long p2 = 37;
const long long p3 = 29;
const int ver1 = (62 * 62);
long long dp[20][11][11][4][4];
int main() {
  int n, t;
  cin >> n >> t;
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < 4; ++j) {
      if (i == j) {
        continue;
      }
      dp[1][0][0][i][j] = 1;
    }
  }
  for (int i = 1; i < n - 1; ++i) {
    for (int j = 0; j <= t; ++j) {
      for (int k = 0; k <= t - 1; ++k) {
        for (int q = 0; q < 4; ++q) {
          for (int w = 0; w < 4; ++w) {
            if (q == w) {
              continue;
            }
            for (int e = 0; e < 4; ++e) {
              if (e == w) {
                continue;
              }
              if (q > w) {
                if (w > e) {
                  dp[i + 1][j][k][w][e] += dp[i][j][k][q][w];
                } else {
                  dp[i + 1][j][k + 1][w][e] += dp[i][j][k][q][w];
                }
              } else {
                if (w > e) {
                  dp[i + 1][j + 1][k][w][e] += dp[i][j][k][q][w];
                } else {
                  dp[i + 1][j][k][w][e] += dp[i][j][k][q][w];
                }
              }
            }
          }
        }
      }
    }
  }
  long long ans = 0;
  for (int q = 0; q < 4; ++q) {
    for (int w = 0; w < 4; ++w) {
      ans += dp[n - 1][t][t - 1][q][w];
    }
  }
  cout << ans;
  return 0;
}
