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[20 + 10][5][5][11][11], n, t;
bool v[20 + 10][5][5][11][11];
int solve(int idx, int pp, int p, int ups, int dwns) {
  if (p < 1 || p > 4 || ups > t || idx > n) return 0;
  if (idx == n && ups == t && dwns == t - 1) return 1;
  if (idx == n) return 0;
  if (v[idx][pp][p][ups][dwns]) return dp[idx][pp][p][ups][dwns];
  if (pp < p)
    dp[idx][pp][p][ups][dwns] += solve(idx + 1, p, p + 1, ups, dwns) +
                                 solve(idx + 1, p, p + 2, ups, dwns) +
                                 solve(idx + 1, p, p + 3, ups, dwns);
  if (pp < p)
    dp[idx][pp][p][ups][dwns] += solve(idx + 1, p, p - 1, ups + 1, dwns) +
                                 solve(idx + 1, p, p - 2, ups + 1, dwns) +
                                 solve(idx + 1, p, p - 3, ups + 1, dwns);
  if (pp > p)
    dp[idx][pp][p][ups][dwns] += solve(idx + 1, p, p + 1, ups, dwns + 1) +
                                 solve(idx + 1, p, p + 2, ups, dwns + 1) +
                                 solve(idx + 1, p, p + 3, ups, dwns + 1);
  if (pp > p)
    dp[idx][pp][p][ups][dwns] += solve(idx + 1, p, p - 1, ups, dwns) +
                                 solve(idx + 1, p, p - 2, ups, dwns) +
                                 solve(idx + 1, p, p - 3, ups, dwns);
  v[idx][pp][p][ups][dwns] = true;
  return dp[idx][pp][p][ups][dwns];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int x = 0;
  cin >> n >> t;
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= 4; j++)
      if (i == j)
        continue;
      else
        x += solve(2, i, j, 0, 0);
  cout << x << endl;
  return 0;
}
