Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box). 

Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.

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Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?

Input

The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).

Output

Output a single integer — the minimal possible number of piles.

Examples

Input

3
0 0 10


Output

2


Input

5
0 1 2 3 4


Output

1


Input

4
0 0 0 0


Output

4


Input

9
0 1 0 2 0 1 1 2 10


Output

3

Note

In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.

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In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).

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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class scratch_28 {
    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /** call this method to initialize reader for InputStream */
        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input) );
            tokenizer = new StringTokenizer("");
        }

        /** get next word */
        static String next() throws IOException {
            while ( ! tokenizer.hasMoreTokens() ) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine() );
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt( next() );
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble( next() );
        }
        static long nextLong() throws IOException{
            return Long.parseLong(next());
        }
    }
    public static void main(String[] args) throws IOException {
        Reader.init(System.in);
        int n= Reader.nextInt();
        int arr[]= new int[101];
        for (int i = 0; i <n ; i++) {
            arr[Reader.nextInt()]+=1;
        }
      //  System.out.println("arr="+Arrays.toString(arr));
       // System.out.println(fxn(1,arr,n));
        int min=1;
        int max=n;
       // System.out.println(fxn(3,arr,n));
        while(max-min>1){
            int mid= min+(max-min)/2;
            if(fxn(mid,arr,n)){
                max=mid;
            }
            else{
                min=mid;
            }



        }

        if(fxn(max-1,arr,n)){
            max-=1;
        }
        if(max==0){
            max=1;
        }
        System.out.println(max);

     //   System.out.println(fxn(3,arr,n));





                    
                    
                    
                    
                    
                    






















    }
    static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        long L[] = new long [n1];
        long R[] = new long [n2];

        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
    static boolean ans(int check, long arr[],int n,int med,int k){
        long ans=0;
        for (int i = med; i <n ; i++) {
            if(check>=arr[i]){
                ans+=check-arr[i];
            }
        }
        return ans<=k;




    }
    static boolean fxn(int mid, int[] arr1, int n){
        // checks if arr can be divided into mid number of arrays
        int arr[]= new int[101];
        for (int i = 0; i <101 ; i++) {
            arr[i]= arr1[i];
        }
        int count=n;
        boolean b =true;
        for (int i = 0; i <n ; i++) {
            int ans=0;
                for (int j = i; j < arr.length; j++) {
                    if (arr[j] > 0) {
                     //   System.out.println(i+" "+j);
                        ans+=arr[j];
                        arr[j]=0;
                        if(ans>=mid){
                            arr[j]+=ans-mid;
                            ans=mid;
                            break;
                        }
                        //count--;
                    }
                }
            count-=ans;
           if(count<=0){
               return true;
           }
           else if(count>0 && ans>=mid){
               b=true;
           }
           else if(count>0 && ans<mid){
               return false;
           }








        }















return b;


    }



}