Problem:
Score : 300 points

Problem Statement

There are N observatories in AtCoder Hill, called Obs. 1, Obs. 2, ..., Obs. N. The elevation of Obs. i is H_i.
There are also M roads, each connecting two different observatories. Road j connects Obs. A_j and Obs. B_j.

Obs. i is said to be good when its elevation is higher than those of all observatories that can be reached from Obs. i using just one road.
Note that Obs. i is also good when no observatory can be reached from Obs. i using just one road.

How many good observatories are there?

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq H_i \leq 10^9

1 \leq A_i,B_i \leq N

A_i \neq B_i

Multiple roads may connect the same pair of observatories.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
H_1 H_2 ... H_N
A_1 B_1
A_2 B_2
:
A_M B_M

Output

Print the number of good observatories.

Sample Input 1

4 3
1 2 3 4
1 3
2 3
2 4

Sample Output 1

2

From Obs. 1, you can reach Obs. 3 using just one road. The elevation of Obs. 1 is not higher than that of Obs. 3, so Obs. 1 is not good.

From Obs. 2, you can reach Obs. 3 and 4 using just one road. The elevation of Obs. 2 is not higher than that of Obs. 3, so Obs. 2 is not good.

From Obs. 3, you can reach Obs. 1 and 2 using just one road. The elevation of Obs. 3 is higher than those of Obs. 1 and 2, so Obs. 3 is good.

From Obs. 4, you can reach Obs. 2 using just one road. The elevation of Obs. 4 is higher than that of Obs. 2, so Obs. 4 is good.

Thus, the good observatories are Obs. 3 and 4, so there are two good observatories.

Sample Input 2

6 5
8 6 9 1 2 1
1 3
4 2
4 3
4 6
4 6

Sample Output 2

3

Solution in C# (.NET Core 3.1.201) (judge verdict: Time Limit Exceeded)

//
// ABC166C
// URL : https://atcoder.jp/contests/abc163/tasks/abc166_c
// 
using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;

namespace AtCoder
{
    class ABC166C
    {
        public static readonly long MOD_CONST = 1000000007;//(long)Math.Pow(10, 9) + 7;


        static void Main(string[] args)
        {
            //出力高速化
            var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);
            Solve();
            Console.Out.Flush();
        }

        public class ViewPoint
        {
            public long H;//標高
            public HashSet<int> VListIdx;//つながっている展望台の配列インデックス.重複を許さない
            public bool isGoodKouho;
            public ViewPoint()
            {
                H = 0;
                VListIdx = null;
                isGoodKouho = true;
            }
        }
        public static void Solve()
        {

            int N, M;
            {
                long[] inputLongArray = Console.ReadLine().Split(' ').Select(_inputStr => long.Parse(_inputStr)).ToArray();
                N = (int)inputLongArray[0];//展望台数
                M = (int)inputLongArray[1];//未知数
            }
            ViewPoint[] VPoints = new ViewPoint[N];

            //標高リスト
            long[] HList = Console.ReadLine().Split(' ').Select(_inputStr => long.Parse(_inputStr)).ToArray();
            for(int i = 0; i < HList.Length; ++i)
            {
                VPoints[i] = new ViewPoint();
                VPoints[i].H = HList[i];
                VPoints[i].VListIdx = new HashSet<int>(M);
                VPoints[i].isGoodKouho = true;//最初はみんないい展望台扱い
            }
            //道
            for (int i = 0; i < M; ++i)
            {
                long[] inputLongArray = Console.ReadLine().Split(' ').Select(_inputStr => long.Parse(_inputStr)).ToArray();
                int va = (int)inputLongArray[0] - 1;//0 originにする
                int vb = (int)inputLongArray[1] - 1;

                VPoints[va].VListIdx.Add(vb);
                VPoints[vb].VListIdx.Add(va);
            }

            long ans = 0;//いい展望台の数

            foreach(ViewPoint v in VPoints)
            {
                long currentH = v.H;
                bool found = true;
                if (v.isGoodKouho == false)
                {
                    continue;//もう候補から外れているので調べない
                }
                else
                {
                    foreach (int vidx in v.VListIdx)
                    {
                        if (currentH <= VPoints[vidx].H)
                        {
                            found = false;
                            v.isGoodKouho = false;//自分よりいい展望台があったので、自分を候補から除外
                            break;
                        }
                        else//低い展望台なので、いい展望台候補からは外せる
                        {
                            VPoints[vidx].isGoodKouho = false;
                        }
                    }
                }
                if (found)
                {
                    //Console.WriteLine(v.H);
                    ans++;
                }
            }


            Console.WriteLine(ans);
        }

        //---------------------------------------------------------------------------------

        /// <summary>
        /// 指定された底log_baseでの, 対数を計算
        /// </summary>
        /// <param name="x"></param>
        /// <param name="log_base">指定の底</param>
        /// <returns></returns>
        public static double LogBase(double x, double log_base)
        {
            //log_b(N) = log_a(N) / log_a(B)より
            return Math.Log(x) / Math.Log(log_base);
        }

        /// <summary>
        /// 初項a, 公差d, 項数nの数列の和
        /// </summary>
        /// <param name="n">求めたいところ</param>
        /// <returns>初項からxまで足した値</returns>
        static long SumSeries(long a, long d, long n)
        {
            return (2 * a + (n-1) * d) * n / 2;
        }
        /// <summary>
        /// 累積和の配列を計算する
        /// Sum[ a[0], a[i] )がs[i+1]に入っている
        /// s[0]  : 0                //一切和を取らない場合
        /// s[1]  : ls[0]            //a[0]   + s[0] = a[0]
        /// s[2]  : ls[1] + s[1]     //a[1]   + a[0]
        /// ...
        /// s[N]  : ls[N-1] + s[N-1] //a[N-1] + ... + a[0]
        /// s[N+1]: ls[N] + s[N]     //a[N]   + ... + a[0]
        /// 
        /// 例: [1, 7)の和(a[1] + ... + a[6])
        /// s[1] = a[0] //[0, 1)
        /// s[7] = a[0] + a[1] + ... + a[6] //[0, 7)
        /// よってs[7] - s[1]);
        /// </summary>
        /// <param name="ls">計算したい配列</param>
        /// <returns>累積和配列</returns>
        static List<long> AggregateSumList(List<long> a)
        {
            var s = new List<long>(a.Count+1);
            if(a.Count == 0)
            {
                return s;
            }
            s.Add(0);//s[0] = 0
            for(int i = 0; i < a.Count; ++i)
            {
                s.Add(a[i] + s[i]);
            }
            return s;
        }


        /// <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 = 1; i <= n; ++i)
            {
                ans = 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
        }
    }
}