Problem:
Score : 1600 points

Problem Statement

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.

Sample Input 1

3 3

Sample Output 1

20

There are 20 ways to give 3 red balls and 3 blue balls.
It turns out that all of them are possible.

Here is an example of the operation (r stands for red, b stands for blue):

You choose s = 3, t = 4.

Initially, the row looks like rrrbbb.

You remove 3rd ball (r) and give it to Snuke. Now the row looks like rrbbb.

You remove 4th ball (b) and give it to Snuke. Now the row looks like rrbb.

You remove 1st ball (r) and give it to Snuke. Now the row looks like rbb.

You remove 3rd ball (b) and give it to Snuke. Now the row looks like rb.

You remove 1st ball (r) and give it to Snuke. Now the row looks like b.

You remove 1st ball (b) and give it to Snuke. Now the row is empty.

This way, Snuke receives balls in the order rbrbrb.

Sample Input 2

4 4

Sample Output 2

67

There are 70 ways to give 4 red balls and 4 blue balls.
Among them, only bbrrbrbr, brbrbrbr, and brrbbrbr are impossible.

Sample Input 3

7 9

Sample Output 3

7772

Sample Input 4

1987 1789

Sample Output 4

456315553

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Diagnostics;
using Debug = System.Diagnostics.Trace;
using SB = System.Text.StringBuilder;
using System.Numerics;
using static System.Math;
using Number = System.Int32;

namespace Program {
	public class Solver {
		Random rnd = new Random();
		public void Solve() {
			var R = ri;
			var B = ri;
			// B が初めてでてきたタイミングで t が確定する
			// その後の B-1 個取り出したとき、 j 回青が出てきた、とする。
			// C(B-1,j) 通り
			// 0 <= R' <= R-(B-1-j), B' = B-1-j 
			// その後 再び B が初めて出てきたタイミングで s が確定する
			// その後の B'-1 個取り出したとき、 k 回青が出てきたとする
			// C(B'-1, k)
			ModInt ans = 0;

			// 1 回目で i、2 回目で j だけ取り除けるようなやつ
			for (int i = 0; i < B; i++) {
				var b = B - 1 - i;
				if (b == 0) {
					var c = coef[B - 1, i];
					c *= coef[R + 1, 1];
					ans += c;
				}
				for (int j = 0; j < b; j++) {
					var c = coef[B - 1, i] * coef[b - 1, j];
					int rem = R - (B - 1 - i) - (b - 1 - j);
					c *= coef[rem + 2, 2];
					ans += c;

				}
			}
			Console.WriteLine(ans);
		}
		BinomialCoefficient coef = new BinomialCoefficient(5050);
		const long INF = 1L << 60;
		static int[] dx = { -1, 0, 1, 0 };
		static int[] dy = { 0, 1, 0, -1 };
		int ri { get { return sc.Integer(); } }
		long rl { get { return sc.Long(); } }
		double rd { get { return sc.Double(); } }
		string rs { get { return sc.Scan(); } }
		public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

		static T[] Enumerate<T>(int n, Func<int, T> f) {
			var a = new T[n];
			for (int i = 0; i < a.Length; ++i) a[i] = f(i);
			return a;
		}
		static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
	}
}

#region main
static class Ex {
	static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
	static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
		return string.Join(st, ie);
	}
	static public void Main() {
		Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = true });
		var solver = new Program.Solver();
		//* 
		var t = new System.Threading.Thread(solver.Solve, 50000000);
		t.Start();
		t.Join();
		//*/
		//solver.Solve();
		Console.Out.Flush();
	}
}
#endregion
#region Ex
namespace Program.IO {
	using System.IO;
	using System.Text;
	using System.Globalization;

	public class Printer : StreamWriter {
		public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
		public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
	}

	public class StreamScanner {
		public StreamScanner(Stream stream) { str = stream; }

		public readonly Stream str;
		private readonly byte[] buf = new byte[1024];
		private int len, ptr;
		public bool isEof = false;
		public bool IsEndOfStream { get { return isEof; } }

		private byte read() {
			if (isEof) return 0;
			if (ptr >= len) {
				ptr = 0;
				if ((len = str.Read(buf, 0, 1024)) <= 0) {
					isEof = true;
					return 0;
				}
			}
			return buf[ptr++];
		}

		public char Char() {
			byte b = 0;
			do b = read(); while ((b < 33 || 126 < b) && !isEof);
			return (char)b;
		}
		public string Scan() {
			var sb = new StringBuilder();
			for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
			return sb.ToString();
		}
		public string ScanLine() {
			var sb = new StringBuilder();
			for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
			return sb.ToString();
		}
		public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
		public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
		public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
	}
}

#endregion


#region ModInt
/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt {
	/// <summary>
	/// 剰余を取る値．
	/// </summary>
	public const long Mod = (int)1e9 + 7;

	/// <summary>
	/// 実際の数値．
	/// </summary>
	public long num;
	/// <summary>
	/// 値が <paramref name="n"/> であるようなインスタンスを構築します．
	/// </summary>
	/// <param name="n">インスタンスが持つ値</param>
	/// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
	public ModInt(long n) { num = n; }
	/// <summary>
	/// このインスタンスの数値を文字列に変換します．
	/// </summary>
	/// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
	public override string ToString() { return num.ToString(); }
	public static ModInt operator +(ModInt l, ModInt r) { l.num += r.num; if (l.num >= Mod) l.num -= Mod; return l; }
	public static ModInt operator -(ModInt l, ModInt r) { l.num -= r.num; if (l.num < 0) l.num += Mod; return l; }
	public static ModInt operator *(ModInt l, ModInt r) { return new ModInt(l.num * r.num % Mod); }
	public static implicit operator ModInt(long n) { n %= Mod; if (n < 0) n += Mod; return new ModInt(n); }

	/// <summary>
	/// 与えられた 2 つの数値からべき剰余を計算します．
	/// </summary>
	/// <param name="v">べき乗の底</param>
	/// <param name="k">べき指数</param>
	/// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
	public static ModInt Pow(ModInt v, long k) { return Pow(v.num, k); }

	/// <summary>
	/// 与えられた 2 つの数値からべき剰余を計算します．
	/// </summary>
	/// <param name="v">べき乗の底</param>
	/// <param name="k">べき指数</param>
	/// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
	public static ModInt Pow(long v, long k) {
		long ret = 1;
		for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
			if ((k & 1) == 1) ret = ret * v % Mod;
		return new ModInt(ret);
	}
	/// <summary>
	/// 与えられた数の逆元を計算します．
	/// </summary>
	/// <param name="v">逆元を取る対象となる数</param>
	/// <returns>逆元となるような値</returns>
	/// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
	public static ModInt Inverse(ModInt v) { return Pow(v, Mod - 2); }
}
#endregion

#region Binomial Coefficient
public class BinomialCoefficient {
	public ModInt[] fact, ifact;
	public BinomialCoefficient(int n) {
		fact = new ModInt[n + 1];
		ifact = new ModInt[n + 1];
		fact[0] = 1;
		for (int i = 1; i <= n; i++)
			fact[i] = fact[i - 1] * i;
		ifact[n] = ModInt.Inverse(fact[n]);
		for (int i = n - 1; i >= 0; i--)
			ifact[i] = ifact[i + 1] * (i + 1);
		ifact[0] = ifact[1];
	}
	public ModInt this[int n, int r] {
		get {
			if (n < 0 || n >= fact.Length || r < 0 || r > n) return 0;
			return fact[n] * ifact[n - r] * ifact[r];
		}
	}
	public ModInt RepeatedCombination(int n, int k) {
		if (k == 0) return 1;
		return this[n + k - 1, k];
	}
}
#endregion