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
// In the name of God

#include <iostream>
#include <algorithm>
#include <fstream>
#include <vector>
#include <deque>
#include <assert.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <stdio.h>
#include <string.h>
#include <utility>
#include <math.h>
#include <bitset>
#include <iomanip>
#include <complex>

using namespace std;

#define rep(i, a, b) for (int i = (a), i##_end_ = (b); i < i##_end_; ++i)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define x first
#define y second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool smin(T &a, const T &b)   { return a > b ? a = b : a;    }
template<typename T> inline bool smax(T &a, const T &b)   { return a < b ? a = b : a;    }

typedef long long LL;

const int N = (int) 4005, mod = (int) 1e9 + 7;
int c[N][N];
int main() {
	for (int i = 0; i < N; ++i)
		for (int j = 0; j <= i; ++j)
			c[i][j] = (j == 0 || i == j) ? 1 : (c[i - 1][j - 1] + c[i - 1][j]) % mod;
	int a, b, res = 0;
	cin >> a >> b;
	for (int bl = 1; bl <= b; ++bl)
		for (int br = 1; br + bl <= b; ++br) {
			int rsz = b - bl;
			int ar = rsz - br;
			int al = b - bl;
			int ways = (LL) c[b - 1][bl - 1] * (br == 0 ? 1 : c[rsz - 1][br - 1]) % mod;
			int rem = a - ar - al;
	//		cout << " hi " << bl << ' ' << br << ' ' << al << ' ' << ar << ' ' << rem << ' ' << ways << endl;
			if (rem < 0) continue;
			int s = 3;
			if (br == 0) s = 2;
			ways = (LL) ways * c[rem + s - 1][s - 1] % mod;
			res = (res + ways) % mod;
		}
	cout << (res + a + 1) % mod << endl;
}





