Greg has a pad. The pad's screen is an n × m rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to n from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to m from left to right.

Greg thinks that the pad's screen displays a cave if the following conditions hold:

  * There is a segment [l, r] (1 ≤ l ≤ r ≤ n), such that each of the rows l, l + 1, ..., r has exactly two black cells and all other rows have only white cells. 
  * There is a row number t (l ≤ t ≤ r), such that for all pairs of rows with numbers i and j (l ≤ i ≤ j ≤ t) the set of columns between the black cells in row i (with the columns where is these black cells) is the subset of the set of columns between the black cells in row j (with the columns where is these black cells). Similarly, for all pairs of rows with numbers i and j (t ≤ i ≤ j ≤ r) the set of columns between the black cells in row j (with the columns where is these black cells) is the subset of the set of columns between the black cells in row i (with the columns where is these black cells). 



Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.

Help Greg.

Input

The first line contains two integers n, m — the pad's screen size (1 ≤ n, m ≤ 2000).

Output

In the single line print the remainder after dividing the answer to the problem by 1000000007 (109 + 7).

Examples

Input

1 1


Output

0


Input

4 4


Output

485


Input

3 5


Output

451
#include <bits/stdc++.h>
long long b[2003][2003];
long long c[2003][2003];
long long c2[2003][2003];
long long d[2003][2003];
int n, m;
long long fac[6010];
long long facinv[6010];
long long mul_inv(long long a, long long b) {
  long long b0 = b, t, q;
  long long x0 = 0, x1 = 1;
  if (b == 1) return 1;
  while (a > 1) {
    q = a / b;
    t = b, b = a % b, a = t;
    t = x0, x0 = x1 - q * x0, x1 = t;
  }
  if (x1 < 0) x1 += b0;
  return x1;
}
long long ncr(int x, int y) {
  long long dab = 1;
  dab *= fac[x];
  dab *= facinv[y];
  dab %= 1000000007;
  dab *= facinv[x - y];
  dab %= 1000000007;
  return dab;
}
int main() {
  int i, j, k;
  scanf("%d%d", &n, &m);
  fac[0] = 1;
  for (i = 1; i <= m + 2 * n; i++) {
    fac[i] = fac[i - 1] * i;
    fac[i] %= 1000000007;
    facinv[i] = mul_inv(fac[i], 1000000007);
  }
  for (j = 0; j <= m; j++) {
    for (i = 0; i <= n; i++) {
      if (i == 0 || j == 0) {
        b[i][j] = 1;
        continue;
      }
      b[i][j] = ncr(j + 2 * i - 1, 2 * i) - ncr(j + 2 * i - 2, 2 * i - 1);
      b[i][j] += b[i - 1][j];
      b[i][j] += 1000000007;
      b[i][j] %= 1000000007;
    }
  }
  for (j = 0; j <= m; j++) {
    for (i = 1; i <= n; i++) {
      if (i == 1)
        c[i][j] = b[i - 1][j];
      else
        c[i][j] = b[i - 1][j] - b[i - 2][j];
      c[i][j] += 1000000007;
      c[i][j] %= 1000000007;
    }
  }
  for (j = 2; j <= m; j++) {
    for (i = 1; i <= n; i++) {
      c2[i][j] = c2[i - 1][j] + c[i][j];
      c2[i][j] %= 1000000007;
    }
  }
  long long dab = 0;
  for (j = 2; j <= m; j++) {
    long long sdab = 0;
    for (i = 0; i < n; i++) {
      sdab += c[i + 1][j] * c2[n - i][j];
      sdab %= 1000000007;
    }
    dab += sdab * (m - j + 1);
    dab %= 1000000007;
  }
  printf("%I64d", dab);
  return 0;
}
