There were n types of swords in the theater basement which had been used during the plays. Moreover there were exactly x swords of each type. y people have broken into the theater basement and each of them has taken exactly z swords of some single type. Note that different people might have taken different types of swords. Note that the values x, y and z are unknown for you.

The next morning the director of the theater discovers the loss. He counts all swords — exactly a_i swords of the i-th type are left untouched.

The director has no clue about the initial number of swords of each type in the basement, the number of people who have broken into the basement and how many swords each of them have taken.

For example, if n=3, a = [3, 12, 6] then one of the possible situations is x=12, y=5 and z=3. Then the first three people took swords of the first type and the other two people took swords of the third type. Note that you don't know values x, y and z beforehand but know values of n and a.

Thus he seeks for your help. Determine the minimum number of people y, which could have broken into the theater basement, and the number of swords z each of them has taken.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^{5}) — the number of types of swords.

The second line of the input contains the sequence a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{9}), where a_i equals to the number of swords of the i-th type, which have remained in the basement after the theft. It is guaranteed that there exists at least one such pair of indices (j, k) that a_j ≠ a_k.

Output

Print two integers y and z — the minimum number of people which could have broken into the basement and the number of swords each of them has taken.

Examples

Input


3
3 12 6


Output


5 3


Input


2
2 9


Output


1 7


Input


7
2 1000000000 4 6 8 4 2


Output


2999999987 2


Input


6
13 52 0 13 26 52


Output


12 13

Note

In the first example the minimum value of y equals to 5, i.e. the minimum number of people who could have broken into the basement, is 5. Each of them has taken 3 swords: three of them have taken 3 swords of the first type, and two others have taken 3 swords of the third type.

In the second example the minimum value of y is 1, i.e. the minimum number of people who could have broken into the basement, equals to 1. He has taken 7 swords of the first type.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static  ArrayList<Pair>adj[];
	
	public static void main(String[] args) throws Exception {
		Scanner sc=new Scanner(System.in);
		PrintWriter out=new PrintWriter(System.out);
		int n=sc.nextInt();
		int a[]=new int[n];
		int max=0;
		for (int i = 0; i < a.length; i++) {
			a[i]=sc.nextInt();
			max=Math.max(max, a[i]);
		}
		long sub=0;
		for (int i = 0; i < a.length; i++) {
			sub+=(max-a[i]);
		}
		long ans=0;
		for (int i = 0; i < a.length; i++) {
			ans=gcd(ans,max-a[i]);
		}
		System.out.println((sub/ans)+" "+ans);
		
		
		
		
		out.flush();
	   
				
		
				
	   
	    
	     
	    
	        
			
				
	}
	
	
	static long mod=(long) (1e9+7);
	static boolean f2=false;
	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r)	//C(p x r) = A(p x q) x (q x r) -- O(p x q x r)
	{
		long[][] C = new long[p][r];
		for(int i = 0; i < p; ++i) {
			for(int j = 0; j < r; ++j) {
				for(int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j]+(a2[i][k]%mod * b[k][j]%mod))%mod;
		 C[i][j]%=mod;
				}
		
			}
		}
		return C;
	}
		static long query[];
	static long last1=0;
	static TreeSet<Long> x1;
	static HashMap<Long,Long> map;
	static long max=0;
	static int first=0;
	
	
	
		
		
		
		

		
	
	static ArrayList<Pair> a1[];
	static int memo1[];
	
	static boolean vis[];
	static TreeSet<Integer> set=new TreeSet<Integer>();
	static class Pair  implements Comparable<Pair>
	{
int x;int y;
		Pair(int n,int c){
			x=n;
			y=c;
		}
		@Override
		public int compareTo(Pair o) {
		if(o.x==x) {
			return o.y-y; 
		}
		else {
			return o.x-x;
		}
		}
		
		
		
	}	
	static long modPow(long ways, long count, long mod)	// O(log e)
	{
		ways %= mod;
		long res = 1;
		while(count > 0)
		{
			if((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res%mod;
	}
	
	
	
	static  int  dx[] = {0,0,0,3,3,3,6,6,6};
	static int dy[] = {0,3,6,0,3,6,0,3,6};
	static int memo[][];
	static int a[];
	static int n;
	static int s;
	static boolean f[];

	static long gcd(long ans,long b) {
		if(b==0) {
			return ans;
		}
		return gcd(b,ans%b);
	}
	static int[] isComposite;
	static int[] valid;
	
	static ArrayList<Integer> primes;
	static ArrayList<Integer> l;
	static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
		l=new ArrayList<>();
		valid=new int[N+1];
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				l.add(i);
				valid[i]=1;
				
				for (int j = i*2; j <=N; j +=i) {	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
						
						
						
				      
					}
			}
		
		for(int i=0 ; i<primes.size() ; i++) {
			for(int j:primes) {
				if(primes.get(i)*j>N) {
					break;
				}
				valid[primes.get(i)*j]=1;
			}
		}
	}
	static char c[];
	static class UnionFind {                                              
		int[] p, rank, setSize;
		int numSets;

		public UnionFind(int N) 
		{
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			for (int i = 0; i < N; i++) {  p[i] = i; setSize[i] = 1; }
		}

		public int findSet(int i) { return p[i] == i ? i : (p[i] = findSet(p[i])); }

		public boolean isSameSet(int i, int j) { return findSet(i) == findSet(j); }

		public void unionSet(int i, int j) 
		{ 
			if (isSameSet(i, j)) 
				return;
			numSets--; 
			int x = findSet(i), y = findSet(j);
			if(rank[x] > rank[y]) { p[y] = x; setSize[x] += setSize[y]; }
			else
			{	p[x] = y; setSize[y] += setSize[x];
				if(rank[x] == rank[y]) rank[y]++; 
			} 
		}

		public int numDisjointSets() { return numSets; }

		public int sizeOfSet(int i) { return setSize[findSet(i)]; }
	}
	public static int[] schuffle(int[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x=(int)(Math.random()*a2.length);
		int temp=a2[x];
		a2[x]=a2[i];
		a2[i]=temp;
		}
		return a2;
	}
		static int V;
	static int INF=(int) 1E9;
		
		static class Edge2 
		{
			int node;
		long cost;
			long next;
			Edge2(int a, int c,Long long1) { node = a;	cost = long1; next=c;}
			
		}
static class Scanner {
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}	
}
