Problem:
Score : 300 points

Problem Statement

Snuke loves puzzles.

Today, he is working on a puzzle using S- and c-shaped pieces.
In this puzzle, you can combine two c-shaped pieces into one S-shaped piece, as shown in the figure below:

Snuke decided to create as many Scc groups as possible by putting together one S-shaped piece and two c-shaped pieces.

Find the maximum number of Scc groups that can be created when Snuke has N S-shaped pieces and M c-shaped pieces.

Constraints

1 ≤ N,M ≤ 10^{12}

Input

The input is given from Standard Input in the following format:

N M

Output

Print the answer.

Sample Input 1

1 6

Sample Output 1

2

Two Scc groups can be created as follows:

Combine two c-shaped pieces into one S-shaped piece

Create two Scc groups, each from one S-shaped piece and two c-shaped pieces

Sample Input 2

12345 678901

Sample Output 2

175897

Solution in C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

//
// ABC055C
// URL : https://atcoder.jp/contests/abc055/tasks/abc055_c
// 
using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;

namespace AtCoder
{
    class ABC055C
    {
        static void Main(string[] args)
        {
            preprocess();
            // 整数配列の入力
            long[] inputLongArray = Console.ReadLine().Split(' ').Select(_inputStr => long.Parse(_inputStr)).ToArray();
            long N = inputLongArray[0];//S型
            long M = inputLongArray[1];//C型

            var ans = makePiece(N, M);

            Console.WriteLine(ans.ToString());
            Console.Out.Flush();
        }

        static long makePiece(long NumS, long NumC)
        {
            //SCC1つを以下の組み合わせで錬成する
            //[1] Sx1, Cx2
            //[2] Cx4
            long total = 0;
            while (true)
            {
                bool existS = NumS - 1 >= 0;
                bool existC = NumC - 2 >= 0;
                if(existS == true && existC == true)
                {
                    NumS--;
                    NumC -= 2;
                    total++;
                }
                else
                {
                    break;
                }
            }
            //残りは[2]パターン。CパーツだけでSCCを構成する.
            total += (NumC / 4);

            return total;
        }

        //---------------------------------------------------------------------------------
        /// <summary>
        /// 
        /// </summary>
        public static void preprocess()
        {
            //出力高速化
            var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);
        }


        /// <summary>
        /// 与えられた数が素数かどうかを計算する
        /// </summary>
        /// <param name="num">対象数</param>
        /// <returns>true:素数, false:素数ではない(合成数)</returns>
        public static bool isPrime(int num)
        {
            if (num < 2) return false;
            else if (num == 2) return true;
            else if (num % 2 == 0) return false;

            double sqrtNum = Math.Sqrt(num);
            for (int i = 3; i <= sqrtNum; i += 2)
            {
                if (num % i == 0)
                {
                    return false;
                }
            }
            return true;
        }

        /// <summary>
        /// 順列mPnを計算する
        /// </summary>
        /// <param name="m"></param>
        /// <param name="n"></param>
        /// <returns></returns>
        public static long Permutation(long m, long n)
        {
            long ret = 1;
            for(var i = 0; i < n; ++i)
            {
                ret *= m - i;
            }
            return ret;
        }
        /// <summary>
        /// 組み合わせmCnを計算する
        /// </summary>
        /// <param name="m"></param>
        /// <param name="n"></param>
        /// <returns></returns>
        public static long Combination(long m, long n)
        {
            if(m == n || n == 0)
            {
                return 1;
            }
            long ans = 1;
            for(var i = 0; i <= n; ++i)
            {
                ans *= (m - i + 1) / i;
            }
            return ans;
        }

        static long Gcd(long m, long n)
        {
            if (n == 0) return m;
            return Gcd(n, m % n);
        }

        static long Lcm(long m, long n)
        {
            var gcd = Gcd(m, n);
            //return (m * n) / gcd;
            return m / gcd * n;//桁あふれ対策
        }
    }
}