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;
template <class T>
T __sqr(const T x) {
  return x * x;
}
template <class T, class X>
inline T __pow(T a, X y) {
  T z = 1;
  for (int i = 1; i <= y; i++) {
    z *= a;
  }
  return z;
}
template <class T>
inline T gcd(T a, T b) {
  a = abs(a);
  b = abs(b);
  if (!b) return a;
  return __gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / __gcd(a, b)) * b;
}
inline bool ispow2(int x) { return (x != 0 && (x & (x - 1)) == 0); }
template <class T>
void UpdateMin(T &x, T y) {
  if (y < x) {
    x = y;
  }
}
template <class T>
void UpdateMax(T &x, T y) {
  if (x < y) {
    x = y;
  }
}
template <class T, class X, class Y>
inline T bigmod(T n, X m, Y mod) {
  unsigned long long ret = 1, a = n % mod;
  while (m) {
    if (m & 1) ret = (ret * a) % mod;
    m >>= 1;
    a = (a * a) % mod;
  }
  ret %= mod;
  return (T)ret;
}
template <class T, class Y>
inline T modinv(T n, Y mod) {
  return bigmod(n, mod - 2, mod);
}
template <class T, class X>
int getbit(T s, X i) {
  return (s >> i) & 1;
}
template <class T, class X>
T onbit(T s, X i) {
  return s | (T(1) << i);
}
template <class T, class X>
T offbit(T s, X i) {
  return s & (~(T(1) << i));
}
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
void extended_euclid(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return;
  }
  long long first, second;
  extended_euclid(b, a % b, first, second);
  x = second;
  y = first - (a / b) * second;
}
pair<long long, pair<long long, long long> > extendedEuclid(long long a,
                                                            long long b) {
  long long x = 1, y = 0;
  long long xLast = 0, yLast = 1;
  long long q, r, m, n;
  while (a != 0) {
    q = b / a;
    r = b % a;
    m = xLast - q * x;
    n = yLast - q * y;
    xLast = x, yLast = y;
    x = m, y = n;
    b = a, a = r;
  }
  return make_pair(b, make_pair(xLast, yLast));
}
const long long mod[] = {0,          1000000007, 1000000009,
                         1000000021, 1000000033, 1000000097,
                         1000000093, 1000000097, 1000000103};
const long long base[] = {0,       1000003, 1000033, 1000037, 1000039,
                          1000081, 1000099, 1000117, 1000121};
int dp[22][12][5][2];
int go(int n, int t, int lst, int up) {
  if (t < 0) return 0;
  if (n == 0) return (t == 0 && up == 0);
  int &ret = dp[n][t][lst][up];
  if (ret == -1) {
    ret = 0;
    for (int i = 1; i <= 4; i++) {
      if (i > lst) {
        ret += go(n - 1, t, i, 1);
      } else if (i < lst) {
        ret += go(n - 1, t - up, i, 0);
      }
    }
  }
  return ret;
}
int main() {
  int n, t;
  cin >> n >> t;
  int ans = 0;
  for (int i = 1; i <= 4; i++) {
    for (int j = i + 1; j <= 4; j++) {
      memset(dp, -1, sizeof dp);
      ans += go(n - 2, t, j, 1);
    }
  }
  cout << ans << endl;
  return 0;
}
