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 <cmath>
#include <cstdio>
#include <cstring>
#include <complex>
#include <algorithm>

using namespace std;
typedef long long LL;
typedef complex<long double> CP;
const int Mod=(1e9)+7;
int n,m,N,M,ans,i,j,k,f[2005][2005],h[2005][2005];
int sum[2005][2005],g[2005][2005],rev[10000],dig[30];
CP a[10000],b[10000];

void FFT(CP a[],int flag){
  for (int i=0;i<N;i++)
    if (i<rev[i]) swap(a[i],a[rev[i]]);
  for (int i=2;i<=N;i<<=1){
    CP wn = CP ( cos(M_PI*2/i), flag*sin(M_PI*2/i) );
    for (int j=0;j<N;j+=i){
      CP w = CP ( 1, 0 );
      for (int k=j;k<j+i/2;k++){
        CP x=a[k], y=a[k+i/2]*w;
        a[k]=x+y; a[k+i/2]=x-y; w=w*wn;
      }
    }
  }
  if (flag==1) return;
  for (int i=0;i<N;i++) a[i].real()/=N;
}

int main(){
  //freopen("295D.in","r",stdin);
  //freopen("295D.out","w",stdout);
  scanf("%d%d",&n,&m);
  
  for (i=2;i<=m;i++){
    f[i][1]=h[i][0]=1;
    sum[i][1]=i-1;
  }
  for (i=2;i<=n;i++)
    for (j=2;j<=m;j++){
      f[j][i]=(f[j-1][i]+sum[j][i-1])%Mod;
      h[j][i-1]=(f[j][i]-f[j][i-1]+Mod)%Mod;
      sum[j][i]=(f[j][i]+sum[j-1][i])%Mod;
    }
  
  for (N=2, M=1;N<((n+1)<<1);N<<=1,M++);
  for (i=0;i<N;i++){
    int len=0;
    for (j=i;j>0;j>>=1) dig[len++]=j&1;
    for (j=0;j<M;j++) rev[i]=( (rev[i]<<1)|dig[j] );
  }
  
  for (i=2;i<=m;i++){
    /*for (j=1;j<=n;j++)
      for (k=0;k<=j;k++)
        g[i][j]=(g[i][j]+(LL)f[i][k]*h[i][j-k]%Mod)%Mod;*/
    for (j=0;j<=n;j++) a[j]=f[i][j];
    for (j=0;j<=n;j++) b[j]=h[i][j];
    for (j=n+1;j<N;j++) a[j]=b[j]=0;
    FFT(a,1); FFT(b,1);
    for (j=0;j<N;j++) a[j]=a[j]*b[j];
    FFT(a,-1);
    for (j=1;j<=n;j++) g[i][j]=( LL (a[j].real()+0.5) )%Mod;
  }
  for (i=1;i<=m;i++)
    for (j=1;j<=n;j++)
      ans=(ans+(LL)g[i][j]*(m-i+1)*(n-j+1)%Mod)%Mod;
  printf("%d\n",ans);
  return 0;
}