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 <bits/stdc++.h>

using namespace std;

const int MX = 4001, md = 1000000007;

int C[MX][MX], S[MX][MX], f[MX][MX];

int main() {
	S[0][0] = C[0][0] = 1;
	for (int i = 1; i < MX; i++) {
		S[i][0] = C[i][0] = 1;
		for (int j = 1; j <= i; j++) {
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % md;
			S[i][j] = (S[i][j - 1] + C[i][j]) % md;
		}
	}

	for (int a = 0; a < MX; a++)
		for (int b = 0; b < MX; b++) {
			if (a == 0 || b == 0) {
				f[a][b] = 1;
				continue;
			}
			
			f[a][b] = (f[a - 1][b] + S[b - 1][min(b - 1, a)]) % md;
		}
	
	int a, b, ans = 0;
	ignore = scanf("%d %d", &a, &b);
	if (a == 0 || b == 0) {
		printf("1\n");
		return 0;
	}
	
	for (int k = 0; k <= a; k++) {
		int left = a - k;
		for (int x = 0; x < b && x <= left; x++)
			ans = (ans + C[b - 1][x] * 1ll * f[left - x][x]) % md;
	}
	
	printf("%d\n", ans);
	
	return 0;
}
