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 N = 2005;
const int mod = 1e9 + 7;

int nr, nb;
int c[N][N];
int g[N][N];
int f[N][N][2];

int cal(int r, int b, int t) {
	int &ret = f[r][b][t];
	if (ret != -1) return ret;
	if (r == nr && b == nb) return 1;
	ret = 0;
	if (!t) {
		int need = nb - 1;
		for (int i = 0; i <= need; ++i) {
			if (need - i + r > nr) continue;
			ret = (ret + 1LL * c[need][i] * cal(r + need - i, 1 + i, 1)) % mod; 
			//cout << "at " << r << ' ' << b << " cal " << r + need - i << ' ' << 1 + i << ' ' << cal(r + need - i, 1 + i, 1) << '\n';
		}
		if (r < nr) ret = (ret + cal(r + 1, 0, 0)) % mod;
		return ret;
	}
	int need = nb - b - 1;
	if (need >= 0) {
		int tmp = max(0, need + r - nr);
		ret = (ret + g[need][need]) % mod;
		if (tmp) {
			ret = (ret - g[need][tmp - 1] + mod) % mod;
		}
	}	
	if (r < nr) ret = (ret + cal(r + 1, b, 1)) % mod;
	return ret;
}

int main() {
	cin >> nr >> nb;
	for (int i = 0; i < N; ++i) g[i][0] = c[i][0] = 1;
	for (int i = 1; i < N; ++i) {
		for (int j = 1; j <= i; ++j) {
			c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
			g[i][j] = (g[i][j - 1] + c[i][j]) % mod;
		}
	}
	memset(f, -1, sizeof f);
	cout << cal(0, 0, 0);
}