A + B balls are arranged in a row. The leftmost A balls are colored red, and the rightmost B balls are colored blue.

You perform the following operation:

* First, you choose two integers s, t such that 1 \leq s, t \leq A + B.
* Then, you repeat the following step A + B times: In each step, you remove the first ball or the s-th ball (if it exists) or the t-th ball (if it exists, all indices are 1-based) from left in the row, and give it to Snuke.



In how many ways can you give the balls to Snuke? Compute the answer modulo 10^9 + 7.

Here, we consider two ways to be different if for some k, the k-th ball given to Snuke has different colors. In particular, the choice of s, t doesn't matter. Also, we don't distinguish two balls of the same color.

Constraints

* 1 \leq A, B \leq 2000

Input

Input is given from Standard Input in the following format:


A B


Output

Print the answer.

Examples

Input

3 3


Output

20


Input

4 4


Output

67


Input

7 9


Output

7772


Input

1987 1789


Output

456315553
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
const ll mod = 1e9+7;
const ll maxn = 2010;
ll a,b,ans,c[maxn][maxn],sum[maxn][maxn],f[maxn][maxn];

void pre()
{
    for (ll i = 0; i < maxn; i++)
    {
        c[i][0] = sum[i][0] = f[i][0] = 1;
        for (ll j = 1; j <= i; j++)
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
        for (ll j = 1; j < maxn; j++)
            sum[i][j] = (sum[i][j - 1] + c[i][j]) % mod;
        for (ll j = 1; j < maxn; j++)
            f[i][j] = (f[i][j - 1] + sum[i][j]) % mod;
    }
}

ll solve(ll t,ll b)
{
    ll res = 1;
    for (ll i = 1; i <= min(t - 1,b); i++)
        res = (res + c[b][i] * f[i - 1][t - i - 1] % mod) % mod;
    return res;
}

int main()
{
    pre();
    scanf("%lld%lld",&a,&b);
    if (!a || !b)
        puts("1");
    else
    {
        for (ll i = 1; i <= a + 1; i++)
            ans = (ans + solve(i,b - 1)) % mod;
        printf("%lld\n",ans);
    }

    return 0;
}
