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>
using namespace std;
template <typename T>
void r1(T &x) {
  x = 0;
  char c(getchar());
  int f(1);
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; '0' <= c && c <= '9'; c = getchar()) x = (x * 10) + (c ^ 48);
  x *= f;
}
const int mod = 1e9 + 7;
template <int mod>
struct typemod {
  int z;
  typemod(int a = 0) : z(a) {}
  inline int inc(int a, int b) const {
    return a += b - mod, a + ((a >> 31) & mod);
  }
  inline int dec(int a, int b) const { return a -= b, a + ((a >> 31) & mod); }
  inline int mul(int a, int b) const { return 1ll * a * b % mod; }
  typemod<mod> operator+(const typemod<mod> &x) const {
    return typemod(inc(z, x.z));
  }
  typemod<mod> operator-(const typemod<mod> &x) const {
    return typemod(dec(z, x.z));
  }
  typemod<mod> operator*(const typemod<mod> &x) const {
    return typemod(mul(z, x.z));
  }
  typemod<mod> &operator+=(const typemod<mod> &x) {
    *this = *this + x;
    return *this;
  }
  typemod<mod> &operator-=(const typemod<mod> &x) {
    *this = *this - x;
    return *this;
  }
  typemod<mod> &operator*=(const typemod<mod> &x) {
    *this = *this * x;
    return *this;
  }
  int operator==(const typemod<mod> &x) const { return x.z == z; }
  int operator!=(const typemod<mod> &x) const { return x.z != z; }
};
template <typename T, typename... Args>
inline void r1(T &t, Args &...args) {
  r1(t);
  r1(args...);
}
const int maxn = 2e3 + 5;
const int maxm = maxn << 1;
typemod<mod> f[maxn][maxn], g[maxn][maxn];
int n, m;
signed main() {
  int i, j;
  r1(n, m);
  for (i = 2; i <= m; ++i) f[1][i] = 1;
  for (i = 2; i <= n; ++i) {
    for (j = 2; j <= m; ++j) {
      f[i][j] = f[i - 1][j] + f[i][j - 1] + f[i][j - 1] - f[i][j - 2];
    }
  }
  for (i = 1; i <= n; ++i) {
    for (j = 2; j <= m; ++j) g[i][j] = g[i - 1][j] + f[i][j];
  }
  typemod<mod> res(0);
  for (i = 1; i <= n; ++i)
    for (j = 2; j <= m; ++j) res += f[i][j] * g[n - i + 1][j] * (m - j + 1);
  printf("%d\n", res.z);
  return 0;
}
