Problem: Vasily the bear has two favorite integers n and k and a pencil. Besides, he's got k jars with different water color paints. All jars are numbered in some manner from 1 to k, inclusive. The jar number i contains the paint of the i-th color. 

Initially the bear took a pencil and drew four segments on the coordinate plane. All of them end at point (0, 0). They begin at: (0, 2n), (0, - 2n), (2n, 0), ( - 2n, 0). Then for each i = 1, 2, ..., n, the bear drew two squares. The first square has the following vertex coordinates: (2i, 0), ( - 2i, 0), (0, - 2i), (0, 2i). The second square has the following vertex coordinates: ( - 2i - 1, - 2i - 1), ( - 2i - 1, 2i - 1), (2i - 1, - 2i - 1), (2i - 1, 2i - 1). After that, the bear drew another square: (1, 0), ( - 1, 0), (0, - 1), (0, 1). All points mentioned above form the set of points A.

<image>

The sample of the final picture at n = 0

<image>

The sample of the final picture at n = 2

The bear decided to paint the resulting picture in k moves. The i-th move consists of the following stages: 

  1. The bear chooses 3 distinct points in set А so that any pair of the chosen points has a segment on the picture between them. The chosen points and segments mark the area that mustn't contain any previously painted points. 
  2. The bear paints the area bounded by the chosen points and segments the i-th color. 



Note that after the k-th move some parts of the picture can stay unpainted.

The bear asked you to calculate, how many distinct ways there are to paint his picture. A way to paint the picture is a sequence of three-element sets of points he chose on each step. Two sequences are considered distinct if there is such number i (1 ≤ i ≤ k), that the i-th members of these sequences do not coincide as sets. As the sought number can be rather large, you only need to calculate the remainder after dividing it by number 1000000007 (109 + 7).

Input

The first line contains two integers n and k, separated by a space (0 ≤ n, k ≤ 200).

Output

Print exactly one integer — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

0 0


Output

1


Input

0 1


Output

8


Input

0 2


Output

32


Input

1 1


Output

32

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int modulo = 1000000007;
long long int combi[10][10];
long long int triangulo1(int n, int k);
long long int triangulo2opuesto(int n, int k);
long long int triangulo2(int n, int k);
long long int triangulo3(int n, int k);
long long int triangulo4(int n, int k);
long long int cuadrado1(int n, int k) {
  long long int mem = 0;
  for (int kk = 0; kk <= 1 and kk <= k; kk++)
    mem = (mem + combi[1][kk] * triangulo1(n - 1, k - kk)) % modulo;
  return mem;
}
long long int cuadrado2opuesto(int n, int k) {
  long long int mem = 0;
  for (int kk = 0; kk <= 2 and kk <= k; kk++)
    mem = (mem + combi[2][kk] * triangulo2opuesto(n - 1, k - kk)) % modulo;
  return mem;
}
long long int cuadrado2(int n, int k) {
  long long int mem = 0;
  for (int kk = 0; kk <= 2 and kk <= k; kk++)
    mem = (mem + combi[2][kk] * triangulo2(n - 1, k - kk)) % modulo;
  return mem;
}
long long int cuadrado3(int n, int k) {
  long long int mem = 0;
  for (int kk = 0; kk <= 3 and kk <= k; kk++)
    mem = (mem + combi[3][kk] * triangulo3(n - 1, k - kk)) % modulo;
  return mem;
}
long long int cuadrado4(int n, int k) {
  long long int mem = 0;
  for (int kk = 0; kk <= 4 and kk <= k; kk++)
    mem = (mem + combi[4][kk] * triangulo4(n - 1, k - kk)) % modulo;
  return mem;
}
long long int memtriangulo1[201][201];
long long int triangulo1(int n, int k) {
  if (k == 0) return 1;
  if (n == 0) {
    if (k == 1) return 1;
    return 0;
  }
  long long int &mem = memtriangulo1[n][k];
  if (mem != -1) return mem;
  mem = 0;
  for (int kk = 0; kk <= 2 and kk <= k; kk++)
    mem = (mem + combi[2][kk] * cuadrado1(n, k - kk)) % modulo;
  if (k == 1) mem = (mem + 1) % modulo;
  return mem;
}
long long int memtriangulo2opuesto[201][201];
long long int triangulo2opuesto(int n, int k) {
  if (k == 0) return 1;
  if (n == 0) {
    if (k == 1) return 2;
    if (k == 2) return 1;
    return 0;
  }
  long long int &mem = memtriangulo2opuesto[n][k];
  if (mem != -1) return mem;
  mem = 0;
  for (int kk = 0; kk <= 4 and kk <= k; kk++)
    mem = (mem + combi[4][kk] * cuadrado2opuesto(n, k - kk)) % modulo;
  for (int kk = 0; kk <= 2 and kk + 1 <= k; kk++)
    mem = (mem + 2 * combi[2][kk] * cuadrado1(n, k - kk - 1)) % modulo;
  if (k == 2) mem = (mem + 1) % modulo;
  return mem;
}
long long int memtriangulo2[201][201];
long long int triangulo2(int n, int k) {
  if (k == 0) return 1;
  if (n == 0) {
    if (k == 1) return 3;
    if (k == 2) return 1;
    return 0;
  }
  long long int &mem = memtriangulo2[n][k];
  if (mem != -1) return mem;
  mem = 0;
  for (int kk1 = 0; kk1 <= 1 and kk1 <= k; kk1++)
    for (int kk2 = 0; kk2 <= 4 - 2 * kk1 and kk2 <= k - kk1; kk2++)
      mem = (mem + combi[1][kk1] * combi[4 - 2 * kk1][kk2] *
                       cuadrado2(n, k - kk1 - kk2)) %
            modulo;
  for (int kk = 0; kk <= 2 and kk + 1 <= k; kk++)
    mem = (mem + 2 * combi[2][kk] * cuadrado1(n, k - kk - 1)) % modulo;
  if (k == 1) mem = (mem + 1) % modulo;
  if (k == 2) mem = (mem + 1) % modulo;
  return mem;
}
long long int memtriangulo3[201][201];
long long int triangulo3(int n, int k) {
  if (k == 0) return 1;
  if (n == 0) {
    if (k == 1) return 5;
    if (k == 2) return 3 + 2;
    if (k == 3) return 1;
    return 0;
  }
  long long int &mem = memtriangulo3[n][k];
  if (mem != -1) return mem;
  mem = 0;
  for (int kk1 = 0; kk1 <= 2 and kk1 <= k; kk1++)
    for (int kk2 = 0; kk2 <= 6 - 2 * kk1 and kk2 <= k - kk1; kk2++)
      mem = (mem + combi[2][kk1] * combi[6 - 2 * kk1][kk2] *
                       cuadrado3(n, k - kk1 - kk2)) %
            modulo;
  for (int kk1 = 0; kk1 <= 1 and kk1 <= k - 1; kk1++)
    for (int kk2 = 0; kk2 <= 4 - 2 * kk1 and kk2 <= k - 1 - kk1; kk2++)
      mem = (mem + 2 * combi[1][kk1] * combi[4 - 2 * kk1][kk2] *
                       cuadrado2(n, k - 1 - kk1 - kk2)) %
            modulo;
  for (int kk = 0; kk <= 4 and kk + 1 <= k; kk++)
    mem = (mem + combi[4][kk] * cuadrado2opuesto(n, k - kk - 1)) % modulo;
  for (int kk = 0; kk <= 2 and kk + 1 <= k; kk++)
    mem = (mem + 2 * combi[2][kk] * cuadrado1(n, k - kk - 1)) % modulo;
  for (int kk = 0; kk <= 2 and kk + 2 <= k; kk++)
    mem = (mem + 3 * combi[2][kk] * cuadrado1(n, k - kk - 2)) % modulo;
  if (k == 2) mem = (mem + 2) % modulo;
  if (k == 3) mem = (mem + 1) % modulo;
  return mem;
}
long long int memtriangulo4[201][201];
long long int triangulo4(int n, int k) {
  if (k == 0) return 1;
  if (n == 0) {
    if (k == 1) return 8;
    if (k == 2) return 6 + 8 + 2;
    if (k == 3) return 4 + 4;
    if (k == 4) return 1;
    return 0;
  }
  long long int &mem = memtriangulo4[n][k];
  if (mem != -1) return mem;
  mem = 0;
  for (int kk1 = 0; kk1 <= 4 and kk1 <= k; kk1++)
    for (int kk2 = 0; kk2 <= 8 - 2 * kk1 and kk2 <= k - kk1; kk2++)
      mem = (mem + combi[4][kk1] * combi[8 - 2 * kk1][kk2] *
                       cuadrado4(n, k - kk1 - kk2)) %
            modulo;
  for (int kk1 = 0; kk1 <= 2 and kk1 <= k - 1; kk1++)
    for (int kk2 = 0; kk2 <= 6 - 2 * kk1 and kk2 <= k - 1 - kk1; kk2++)
      mem = (mem + 4 * combi[2][kk1] * combi[6 - 2 * kk1][kk2] *
                       cuadrado3(n, k - 1 - kk1 - kk2)) %
            modulo;
  for (int kk1 = 0; kk1 <= 1 and kk1 <= k - 2; kk1++)
    for (int kk2 = 0; kk2 <= 4 - 2 * kk1 and kk2 <= k - 2 - kk1; kk2++)
      mem = (mem + 4 * combi[1][kk1] * combi[4 - 2 * kk1][kk2] *
                       cuadrado2(n, k - 2 - kk1 - kk2)) %
            modulo;
  for (int kk = 0; kk <= 4 and kk <= k - 2; kk++)
    mem = (mem + 2 * combi[4][kk] * cuadrado2opuesto(n, k - 2 - kk)) % modulo;
  for (int kk1 = 0; kk1 <= 1 and kk1 <= k - 1; kk1++)
    for (int kk2 = 0; kk2 <= 4 - 2 * kk1 and kk2 <= k - 1 - kk1; kk2++)
      mem = (mem + 4 * combi[1][kk1] * combi[4 - 2 * kk1][kk2] *
                       cuadrado2(n, k - 1 - kk1 - kk2)) %
            modulo;
  for (int kk = 0; kk <= 2 and kk + 2 <= k; kk++)
    mem = (mem + 8 * combi[2][kk] * cuadrado1(n, k - kk - 2)) % modulo;
  for (int kk = 0; kk <= 2 and kk + 3 <= k; kk++)
    mem = (mem + 4 * combi[2][kk] * cuadrado1(n, k - kk - 3)) % modulo;
  if (k == 2) mem = (mem + 2) % modulo;
  if (k == 3) mem = (mem + 4) % modulo;
  if (k == 4) mem = (mem + 1) % modulo;
  return mem;
}
int main() {
  {
    for (int n = 0; n < 10; n++)
      for (int k = 0; k <= n; k++)
        if (k == 0 or k == n)
          combi[n][k] = 1;
        else
          combi[n][k] = (combi[n - 1][k] + combi[n - 1][k - 1]) % modulo;
  }
  for (int i = 0; i < 201; i++)
    for (int j = 0; j < 201; j++)
      memtriangulo1[i][j] = memtriangulo2opuesto[i][j] = memtriangulo2[i][j] =
          memtriangulo3[i][j] = memtriangulo4[i][j] = -1;
  int n, k;
  cin >> n >> k;
  long long int sol = triangulo4(n, k);
  for (int i = 1; i <= k; i++) sol = (sol * i) % modulo;
  cout << sol << endl;
}