Problem:
Score : 300 points

Problem Statement

You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N).
Your objective is to remove some of the elements in a so that a will be a good sequence.

Here, an sequence b is a good sequence when the following condition holds true:

For each element x in b, the value x occurs exactly x times in b.

For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not.

Find the minimum number of elements that needs to be removed so that a will be a good sequence.

Constraints

1 \leq N \leq 10^5

a_i is an integer.

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of elements that needs to be removed so that a will be a good sequence.

Sample Input 1

4
3 3 3 3

Sample Output 1

1

We can, for example, remove one occurrence of 3. Then, (3, 3, 3) is a good sequence.

Sample Input 2

5
2 4 1 4 2

Sample Output 2

2

We can, for example, remove two occurrences of 4. Then, (2, 1, 2) is a good sequence.

Sample Input 3

6
1 2 2 3 3 3

Sample Output 3

0

Sample Input 4

1
1000000000

Sample Output 4

1

Remove one occurrence of 10^9. Then, () is a good sequence.

Sample Input 5

8
2 7 1 8 2 8 1 8

Sample Output 5

5

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import static java.lang.System.out;
import java.util.Scanner;
import java.util.Arrays;

public class Main{
   static Scanner sc = new Scanner(System.in);
   public static void main(String[] args){
      int N = sc.nextInt();
      if(N==0){
         System.out.println(N);
         return;
      }

      int[] array = new int[N];
      for(int i=0; i<N;i++){
         array[i] = sc.nextInt();
      }
      
      Arrays.sort(array);

      int result = 0;
      int count = 0;
      int old = 0;
      //iをグローバルにする？
      for(int i=0;i<N;i++){
         int target = array[i];
         if(target>N){
            result += N-i + count;
            break;
         }
         
         if(old == target){
            count++;

         }else{
            if(old >count){
               result+= count;
            }else{
               result+= count-old;
            }
            old = target;
            count = 1;
         }
         //count と old に

      }
      if(!(old ==count)){
         result += count;
      }
      System.out.println(result);
      /**
       * 
       * 1から順に探索
       * まず前回と同じか見る。
       * 
       * 前回とおなじ
       *      カウント数の方が大きい
       *      カウント数 - ターゲット
       *      カウント数の方が小さい
       *       コンテニュー
       * 
       * 古数とちがう
       *    カウント数加算。
       * 
       *    2 3 
       * 
       * 終了時
       * ばらが残ってる
       * oldとcountが不一致の場合追加
       * 
       * 10回回転
       * 全部11とか
       * result = N-i +count;
       * 
       */

       
   }
}