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 <iostream>

using namespace std;

const int MOD = 1000 * 1000 * 1000 + 7;
const int MAXA = 2010;

int c[MAXA][MAXA];

int add(long long a, long long b)
{
	return (a + b) % MOD;
}

int mul(long long a, long long b)
{
	return (a * b) % MOD;
}

int main()
{
	int a, b;
	cin >> a >> b;
	int ans = 0;
	for (int i = 0; i <= b; i++)
	{
		c[i][0] = c[i][i] = 1;
		for (int j = 1; j < i; j++)
			c[i][j] = add(c[i - 1][j], c[i - 1][j - 1]);
	}
	for (int i = 0; i <= b - 1 && i <= a; i++)
	{
		if (i == 0)
			ans = add(ans, add(a, 1));
		else
			for (int j = 0; j <= i - 1 && (i + j) <= a; j++)
				ans = add(ans, mul(mul(c[b - 1][i], c[i - 1][j]), (a + 1 - i - j) * (a + 2 - i - j) / 2));
	}
	cout << ans << endl;
}