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 t, n, dp[4 + 2][4 + 2][10 + 5][10 + 5][20 + 5];
int solve(int prev, int me, int high, int low, int pos) {
  if (high == 0 && low == 0 && pos == n) {
    return 1;
  }
  if (pos == n || high < 0 || low < 0) return 0;
  if (dp[prev][me][high][low][pos] != -1) return dp[prev][me][high][low][pos];
  int ans = 0;
  for (int i = 1; i <= 4; i++) {
    if (i == me) continue;
    if (i > me) {
      if (prev > me) {
        ans += solve(me, i, high, low - 1, pos + 1);
      } else {
        ans += solve(me, i, high, low, pos + 1);
      }
    } else if (i < me) {
      if (prev < me) {
        ans += solve(me, i, high - 1, low, pos + 1);
      } else {
        ans += solve(me, i, high, low, pos + 1);
      }
    } else {
      ans += solve(me, i, high, low, pos + 1);
    }
  }
  return dp[prev][me][high][low][pos] = ans;
}
int main() {
  cin >> n >> t;
  memset(dp, -1, sizeof dp);
  int ans = 0;
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= 4; j++) {
      if (i == j) continue;
      ans += solve(i, j, t, t - 1, 2);
    }
  }
  cout << ans << endl;
}
