Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, Vlad immediately bought only to find out his computer is too old for the new game and needs to be updated.

There are n video cards in the shop, the power of the i-th video card is equal to integer value ai. As Vlad wants to be sure the new game will work he wants to buy not one, but several video cards and unite their powers using the cutting-edge technology. To use this technology one of the cards is chosen as the leading one and other video cards are attached to it as secondary. For this new technology to work it's required that the power of each of the secondary video cards is divisible by the power of the leading video card. In order to achieve that the power of any secondary video card can be reduced to any integer value less or equal than the current power. However, the power of the leading video card should remain unchanged, i.e. it can't be reduced.

Vlad has an infinite amount of money so he can buy any set of video cards. Help him determine which video cards he should buy such that after picking the leading video card and may be reducing some powers of others to make them work together he will get the maximum total value of video power.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of video cards in the shop.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 200 000) — powers of video cards.

Output

The only line of the output should contain one integer value — the maximum possible total power of video cards working together.

Examples

Input

4
3 2 15 9


Output

27


Input

4
8 2 2 7


Output

18

Note

In the first sample, it would be optimal to buy video cards with powers 3, 15 and 9. The video card with power 3 should be chosen as the leading one and all other video cards will be compatible with it. Thus, the total power would be 3 + 15 + 9 = 27. If he buys all the video cards and pick the one with the power 2 as the leading, the powers of all other video cards should be reduced by 1, thus the total power would be 2 + 2 + 14 + 8 = 26, that is less than 27. Please note, that it's not allowed to reduce the power of the leading video card, i.e. one can't get the total power 3 + 1 + 15 + 9 = 28.

In the second sample, the optimal answer is to buy all video cards and pick the one with the power 2 as the leading. The video card with the power 7 needs it power to be reduced down to 6. The total power would be 8 + 2 + 2 + 6 = 18.
/**
 p * @author prakhar28
 *
 */
import java.io.*;
import java.util.*;
import java.util.Map.Entry;
import java.text.*;
import java.math.*;
import java.util.regex.*;


import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;

public class Main{
	
	static class Reader
    {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public String readLine() throws IOException
        {
            byte[] buf = new byte[200005]; // line length+1
            int cnt = 0, c;
            while ((c = read()) != -1)
            {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }
 
        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
            {
                ret = ret * 10 + c - '0';
            }  while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
 
        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
 
        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
 
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
 
            if (c == '.')
            {
                while ((c = read()) >= '0' && c <= '9')
                {
                    ret += (c - '0') / (div *= 10);
                }
            }
 
            if (neg)
                return -ret;
            return ret;
        }
 
        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
 
        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
 
        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
	
	static class FastScanner {

	    private final BufferedReader bufferedReader;
	    private StringTokenizer stringTokenizer;

	    public FastScanner() {
	        bufferedReader = new BufferedReader(new InputStreamReader(System.in));
	    }

	    public String next() {
	        while (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
	            try {
	                stringTokenizer = new StringTokenizer(bufferedReader.readLine());
	            } catch (IOException e) {
	                throw new RuntimeException("Can't read next value", e);
	            }
	        }
	        return stringTokenizer.nextToken();
	    }

	    public int nextInt() {
	        return Integer.parseInt(next());
	    }

	    public long nextLong() {
	        return Long.parseLong(next());
	    }

	    public double nextDouble() {
	        return Double.parseDouble(next());
	    }

	    public String nextLine(){
	        String str = "";
	        try {
	            str = bufferedReader.readLine();
	        } catch (IOException e) {
	            e.printStackTrace();
	        }
	        return str;
	    }
	}
	
	static void closeall() throws IOException{
		printWriter.close();
		sc.close();
		in.close();
	}
	static Scanner sc = new Scanner(System.in);
	static Reader in = new Reader();
	static FastScanner fastScanner = new FastScanner();
	static PrintWriter printWriter = new PrintWriter(new BufferedOutputStream(System.out));
	static BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
	static long INF = (long)(1e18);

	
	int V,E;
	ArrayList<Integer> adj[];
	HashMap<Integer,Integer> path;
	Main(){
	}
	Main(int v,int e){
		V=v;
		E=e;
		adj=new ArrayList[v];
		path = new HashMap<Integer,Integer>();
		for(int i=0;i<v;i++)
			adj[i] = new ArrayList<>();
	}
	void addEdge(int u,int v){
		adj[u].add(v);
		adj[v].add(u);
	}
	
	
	static long power(long a,long k)
	{
		long m = 1000000007;
	    long ans=1;
	    long tmp=a%m;
	    while(k>0)
	    {
	        if(k%2==1)
	        	ans=ans*tmp%m;
	        tmp=tmp*tmp%m;
	        k>>=1;
	    }
	    return ans;
	}
	
	static class Pair /*implements Comparable<Pair>*/{
		long F,S;
		Pair(long x,long y){
			F = x;
			S = y;
		}
		/*public int compareTo(Pair ob){
			return this.num-ob.num;
		}*/
		
	}
	static long gcd(long a,long b) {
		if(a<b) {
			long t = a;
			a = b;
			b = t;
		}
		long r = a%b;
		if(r==0)
			return b;
		return gcd(b,r);
	}
	
	static int lower_bound(int[] a,int low,int high,int val) {
		while(low!=high) {
			int mid = (low+high)/2;
			if(a[mid]<val)
				low=mid+1;
			else
				high=mid;
		}
		return low;
	}
	
	static int max(int a,int b) {
		return (int)Math.max(a, b);
	}
	static long max(long a,long b) {
		return (long)Math.max(a, b);
	}
	static long min(long a,long b) {
		if(a<b)
			return a;
		return b;
	}
	static long mod = 1000000007;
	
	public static void main(String args[])throws IOException{
		Reader in = new Reader();
		int n = in.nextInt();
		int[] a = new int[n];
		HashSet<Integer> set = new HashSet<>();
		for(int i=0;i<n;i++) {
			a[i] = in.nextInt();
			set.add(a[i]);
		}
		Arrays.sort(a);
		int cnt[] = new int[400005];
		int point = 0;
		for(int i=1;i<400005;i++) {
			while(point<n && a[point]==i) {
				cnt[i]++;
				point++;
			}
			cnt[i]+=cnt[i-1];
		}
		//cnt[i] stores the number of elements less than or equal to i.
		Iterator itr = set.iterator();
		long max = Long.MIN_VALUE;
		while(itr.hasNext()) {
			int num = (int)itr.next();
			long s = 0;
			//elements will be num,2*num,3*num,...
			//number of elements in [j*num,(j+1)*num) are cnt[(j+1)*num-1]-cnt[j*num-1]
			for(int j=1;j*num<=200000;j++) 
				s+=(long)(cnt[(j+1)*num-1]-cnt[j*num-1])*(long)(j*num);
			max = max(max,s);
		}
		printWriter.println(max);
		closeall();
	}
} 