We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller. (The operation is the same as the one in Problem D.)

* Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.



It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.

You are given the sequence a_i. Find the number of times we will perform the above operation.

Constraints

* 2 ≤ N ≤ 50
* 0 ≤ a_i ≤ 10^{16} + 1000

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the number of times the operation will be performed.

Examples

Input

4
3 3 3 3


Output

0


Input

3
1 0 3


Output

1


Input

2
2 2


Output

2


Input

7
27 0 0 0 0 0 0


Output

3


Input

10
1000 193 256 777 0 1 1192 1234567891011 48 425


Output

1234567894848
import java.util.*;
import java.io.*;
import java.math.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class Main{
    static long mod=1000000007;
    // static int dx[]={1,-1,0,0};
    // static int dy[]={0,0,1,-1};
    // static int dx[]={1,-1,0,0,1,1,-1,-1};
    // static int dy[]={0,0,1,-1,1,-1,1,-1};
    // PriorityQueue<Integer> que = new PriorityQueue<Integer>(); 
    //HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
    //ArrayList<Integer> lis = new ArrayList<Integer>();
    public  static void main(String[] args)   throws Exception, IOException{
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);

        int n=sc.nextInt(),a=0;
        long ans=0,e[]=new long[n];
        for( int i=0; i<n; i++ ){
            e[i] = sc.nextLong();
        }
        while(true){
            sort(e);
            int x=-1;
                if(e[n-1]<n) break;
            long c=e[n-1]/n;
            //db(c,x,e,++a);
            ans+=c;
            for( int i=0; i<n-1; i++ ){
                e[i] += c;
            }
            e[n-1] -= c * n;
        }
        out.println(ans); 
        out.flush();
    }

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }

    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }
     
    static boolean bit(long x,int k){
        // weather k-th bit (from right) be one or zero
        return  ( 0 < ( (x>>k) & 1 )  )  ? true:false;
    }
}

class Pair implements Comparable<Pair>{
    int x,y,id;
    Pair(int x, int y,int id) {
        this.x=x;
        this.y=y;
        this.id=id;
    } 
    public int compareTo(Pair p){
        return p.x -     x;
    } 
}

// class Pair implements Comparable<Pair>{
//     int x,y;
//     Pair(int x, int y) {
//         this.x=x;
//         this.y=y;
//     } 
//     public int compareTo(Pair p){//descend
//         return p.x - x;
//     } 

// }

class P implements Comparable<P>{
    int diff,id1,id2;
    P(int diff, int id1,int id2) {
        this.diff=diff;
        this.id1=id1;
        this.id2=id2;
    } 
      
    public int compareTo(P p){//ascend
        return diff - p.diff;
    } 
}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString() throws IOException
    {
        while( st==null || !st.hasMoreTokens() )
            st = new StringTokenizer(x.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException
    {
        return Integer.parseInt(nextString());
    }
    public long nextLong() throws IOException
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}