# include <iostream>
# include <string.h>
# include <algorithm>
# include <math.h>
# include <vector>
# include <stdio.h>
using namespace std;

const int MOD = 1e9+7;

const int MAXN = 1000000;
bool arr[MAXN+100] = {false};

vector<int> produce_prim_number()
{
    vector<int> prim;
    prim.push_back(2);
    int i,j;
    for(i=3; i*i<=MAXN; i+=2)
    {
        if(!arr[i])
        {
            prim.push_back(i);
            for(j=i*i; j<=MAXN; j+=i)
            arr[j] = true;
        }
    }
    while(i<=MAXN)
    {
        if(!arr[i])
        prim.push_back(i);
        i+=2;
    }
    return prim;
}

//计算n!中素因子p的指数
int Cal(int x, int p)
{
    int ans = 0;
    long long rec = p;
    while(x>=rec)
    {
        ans += x/rec;
        rec *= p;
    }
    return ans;
}

//计算n的k次方对M取模，二分法
int Pow(long long n, int k, int MOD)
{
    long long ans = 1;
    while(k)
    {
        if(k&1)
        {
            ans = (ans * n) % MOD;
        }
        n = (n * n) % MOD;
        k >>= 1;
    }
    return ans;
}

//计算C(n,m)
int Combination(int n, int m)
{
    vector<int> prim = produce_prim_number();
    long long ans = 1;
    int num;
    for(int i=0; i<prim.size() && prim[i]<=n; ++i)
    {
        num = Cal(n, prim[i]) - Cal(m, prim[i]) - Cal(n-m, prim[i]);
        ans = (ans * Pow(prim[i], num, MOD)) % MOD;
    }
    return ans;
}

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        n=m+n-2;
        
        printf("%d\n",Combination(n-2,m-2));
    }
    return 0;
}

# include <iostream>
# include <cstdio>
# include <algorithm>
# include <cmath>
# include <cstring>
using namespace std;

# define lld  __int64

lld  n, m, p;

lld Ext_gcd(lld a,lld b,lld &x,lld &y)
{
   if(b==0) { x=1, y=0; return a; }
   lld ret= Ext_gcd(b,a%b,y,x);
   y-= a/b*x;
   return ret;
}

lld Inv(lld a,int m)   ///求逆元
{
   lld d,x,y,t= (lld)m;
   d= Ext_gcd(a,t,x,y);
   if(d==1) return (x%t+t)%t;
   return -1;
}

lld Cm(lld n, lld m, lld p)  ///组合数学
{
    lld a=1, b=1;
    if(m>n) return 0;
    while(m)
    {
        a=(a*n)%p;
        b=(b*m)%p;
        m--;
        n--;
    }
    return (lld)a*Inv(b,p)%p;  ///（a/b）%p 等价于 a*（b，p）的逆元
}

int Lucas(lld n, lld m, lld p)  ///把n分段递归求解相乘
{
    if(m==0) return 1;
    return (lld)Cm(n%p,m%p,p)*(lld)Lucas(n/p,m/p,p)%p;
}

int main()
{
    int  T;
    while(~scanf("%I64d%I64d",&n,&m))
    {
    	n--,m--;
        m--;
        __int64 p=1000000007;
        
        printf("%d\n",Lucas(n+m-1,m,p));
    }
    return 0;
}