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.util.*;
import static java.lang.Math.sqrt;
import static java.lang.Math.round;
import java.lang.Math;
public class Jo {
	public static void main(String args[]) throws IOException 
    { 
		Scanner sc=new Scanner(System.in);
		//System.out.println(largestDivisor(52));
		
		int n=sc.nextInt();long a[]=new long[n];
		for(int i=0;i<n;i++)a[i]=sc.nextInt();Arrays.sort(a);
		long s=0;int p=0;long max=a[a.length-1];
		//System.out.println(findGCD(a, n));
		for(int i=0;i<a.length;i++)s+=(max-a[i]);
		if(findGCD(a, n)!=1)System.out.println(s/findGCD(a, n)+" "+findGCD(a, n));
		else System.out.println(findGCD(a, n)+" "+s/findGCD(a, n));
		
		
	 }
	static long gcd(long a, long a2) 
    { 
        if (a == 0) 
            return a2; 
        return gcd(a2 % a, a); 
    } 
  
    // Function to find gcd of array of 
    // numbers 
    static long findGCD(long[] a, int n) 
    { 
        long result = a[0]; 
        for (int i = 1; i < n; i++) 
            result = gcd(a[i], result); 
  
        return result; 
    } 
	public static int largestDivisor(int n){
		   int divisor = 0;
		   for (int i = 1; i <= n/2; i++)
		       if (n % i == 0) 
		           divisor = i;

		   return divisor; 
		}
	//find the GCD of two numbers in O(min(a,b))
	  static int GCD1(int A, int B) {
		    int m = Math.min(A, B), gcd = 0;
		    for(int i = m; i > 0; --i)
		        if(A % i == 0 && B % i == 0) {
		            gcd = i;
		            break;
		        }
		    return gcd;
		}
	  
	static long sort(int a[],int n)
    {   int b[]=new int[n]; 
        return mergeSort(a,b,0,n-1);}
     
     
    static long mergeSort(int a[],int b[],long left,long right)
    { long c=0;if(left<right)
     {   long mid=left+(right-left)/2;
         c= mergeSort(a,b,left,mid);
         c+=mergeSort(a,b,mid+1,right);
         c+=merge(a,b,left,mid+1,right); }  
        return c;    }
     
     
    static long merge(int a[],int  b[],long left,long mid,long right)
    {long c=0;int i=(int)left;int j=(int)mid; int k=(int)left;
    while(i<=(int)mid-1&&j<=(int)right)
    { if(a[i]<=a[j]) {b[k++]=a[i++]; }
    else    { b[k++]=a[j++];c+=mid-i;}}
    while (i <= (int)mid - 1)   b[k++] = a[i++]; 
    while (j <= (int)right) b[k++] = a[j++];
    for (i=(int)left; i <= (int)right; i++) 
        a[i] = b[i];  return c;  }
	 
	static boolean isPerfectSquare(long x)  
    { 
          
        // Find floating point value of 
        // square root of x. 
        double sr = Math.sqrt(x); 
      
        // If square root is an integer 
        return ((sr - Math.floor(sr)) == 0); 
    } 
    	
	//to check that the string contains the substring      (frequency>=1)
	 static boolean danial(String s) {return s.contains("Danil");}
	 static boolean olya(String s) {return s.contains("Olya");}
	 static boolean Ann(String s) {return s.contains("Ann");}
	 static boolean Slava(String s) {return s.contains("Slava");}
	 static boolean Nikita(String s) {return s.contains("Nikita");}
	 
	 //to check that the string contains the substring only once   (frequency==1)
	  static boolean containsOnce( String s, String k) {
		     //String substring0 = substri.toString();
		     int i = s.indexOf(k);
		    return i != -1 && i == s.lastIndexOf(k);
		}
		 
	  //find the x to the power of n       O(n)
	  int iterativePower(int x,int n)
	  {
	      int result=1;
	      while(n>0)
	      {
	          result=result*x;
	          n--;
	      }
	      return result;
	  }
	  //find x to the power of n         O(log(n))
	  //if (n%2==1).......x^n=x*x^(n-1).....(n-1)is even
	  //if(n%2==0)........x^n=(x^2)^n/2
	  static int binaryExponentiation(int x,int n)
	  {
	      int result=1;
	      while(n>0)
	      {
	    	  //if the given power is odd,it is the start
	    	  //if the given power is even,it is the end
	          if(n % 2 ==1)
	        	  //take one of x to make the power even   ((n-1)%2==0)
	              result=result * x;
	          //whether the power is odd or even
	          x=x*x;
	          n=n/2;
	      }
	      return result;
	  }
	 
	  
	  //find the GCD of two numbers in O(min(a,b))
	  int GCD(int A, int B) {
		    int m = Math.min(A, B), gcd = 0;
		    for(int i = m; i > 0; --i)
		        if(A % i == 0 && B % i == 0) {
		            gcd = i;
		            break;
		        }
		    return gcd;
		}
	  
	  //find the GCD of two numbers in O(log(max(A, B))).
	  //Euclid's algorithm.
	  int GCD2(int A, int B) {
		    if(B==0)
		        return A;
		    else
		        return GCD1(B, A % B);
		}
	
	//count number of digits of number n
	static long countDigit(long n) 
    { 
        return (long)Math.floor(Math.log10(n) + 1); 
    } 
    //min. summation of pairs of divisors
	public static int di(int n) {
		int s;int min=1000000;
		for(int i=1;i*i<=n;i++) {
			if(n%i==0) {
				s=i+(n/i);
				if(s<=min)min=s;
			}
		}
		return min;
	}
	
	public static double squareRoot(int number) {
		double temp;

		double sr = number / 2;

		do {
			temp = sr;
			sr = (temp + (number / temp)) / 2;
		} while ((temp - sr) != 0);

		return sr;
	    }
	static boolean isPerfect(int n) 
    { 
        // To store sum of divisors 
        int sum = 1; 
      
        // Find all divisors and add them 
        for (int i = 2; i * i <= n; i++) 
        { 
            if (n % i==0) 
            { 
                if(i * i != n) 
                    sum = sum + i + n / i; 
                else
                    sum = sum + i; 
            } 
        }  
        // If sum of divisors is equal to 
        // n, then n is a perfect number 
        if (sum == n && n != 1) 
            return true; 
      
        return false; 
    } 
}