Problem:
Score : 400 points

Problem Statement

There is a knight - the chess piece - at the origin (0, 0) of a two-dimensional grid.

When the knight is at the square (i, j), it can be moved to either (i+1,j+2) or (i+2, j+1).

In how many ways can the knight reach the square (X, Y)?

Find the number of ways modulo 10^9 + 7.

Constraints

1 \leq X \leq 10^6

1 \leq Y \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the number of ways for the knight to reach (X, Y) from (0, 0), modulo 10^9 + 7.

Sample Input 1

3 3

Sample Output 1

2

There are two ways: (0,0) \to (1,2) \to (3,3) and (0,0) \to (2,1) \to (3,3).

Sample Input 2

2 2

Sample Output 2

0

The knight cannot reach (2,2).

Sample Input 3

999999 999999

Sample Output 3

151840682

Print the number of ways modulo 10^9 + 7.

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

//author: camypaper
using System;
using System.Linq;
using System.Collections.Generic;
using System.Text.RegularExpressions;
using Debug = System.Diagnostics.Debug;
using SB = System.Text.StringBuilder;
using System.Numerics;
using Point = System.Numerics.Complex;
using static System.Math;
using Number = System.Int64;
using System.Runtime.CompilerServices;

namespace Program {
    public class Solver {
        Random rnd = new Random(0);
        public void Solve() {
            var n = ri;
            var m = ri;
            var k = (n + m) / 3;
            var table = new BinomialCoefficient((n + m + 5));
            n -= k;
            m -= k;
            ModInt ans = table[k, n];
            if (n + m != k || n < 0 || m < 0) ans = 0;
            Console.WriteLine(ans);

        }
        const long INF = 1L << 60;
        int ri => sc.Integer();
        long rl => sc.Long();
        double rd => sc.Double();
        string rs => sc.Scan();
        public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

        static IEnumerable<int> Rep(int n) => Enumerable.Range(0, n);
        static IEnumerable<int> RRep(int n) => Enumerable.Range(0, n).Reverse();
        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 = false });
        var solver = new Program.Solver();
        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 => 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;

        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