Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;
import java.lang.*;
import java.math.BigInteger;
import java.io.*;
 
public class Main {
	static class FastReader { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next() { 
            while (st == null || !st.hasMoreElements()) { 
                try { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() { 
            String str = ""; 
            try { 
                str = br.readLine(); 
            } 
            catch (IOException e) { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
	static int[] cSort(int arr[]) {
		// change the size
		int size = 231;
		
        final int N = arr.length; 
        int output[] = new int[N];         
        int count[] = new int[size]; 
        for (int i=0; i < N; ++i) 
            ++count[arr[i]];
        for (int i=1; i<=size-1; ++i) 
            count[i] += count[i-1];
        for (int i = N-1; i>=0; i--) { 
            output[count[arr[i]]-1] = arr[i]; 
            --count[arr[i]]; 
        }
        return output;
    }
	
	static int RUN = 32;   
    // this function sorts array from left index to  
    // to right index which is of size atmost RUN  
    public static void insertionSort(int[] arr, int left, int right) { 
        for (int i = left + 1; i <= right; i++) { 
            int temp = arr[i]; 
            int j = i - 1; 
            while (j >= left && arr[j] > temp) { 
                arr[j + 1] = arr[j]; 
                j--; 
            } 
            arr[j + 1] = temp; 
        } 
    }
    // merge function merges the sorted runs  
    public static void merge(int[] arr, int l, int m, int r) { 
        // original array is broken in two parts  
        // left and right array  
        int len1 = m - l + 1, len2 = r - m; 
        int[] left = new int[len1]; 
        int[] right = new int[len2]; 
        for (int x = 0; x < len1; x++) { 
            left[x] = arr[l + x]; 
        } 
        for (int x = 0; x < len2; x++) { 
            right[x] = arr[m + 1 + x]; 
        } 
  
        int i = 0; 
        int j = 0; 
        int k = l; 
  
        // after comparing, we merge those two array  
        // in larger sub array  
        while (i < len1 && j < len2) { 
            if (left[i] <= right[j]) { 
                arr[k] = left[i]; 
                i++; 
            } else { 
                arr[k] = right[j]; 
                j++; 
            } 
            k++; 
        } 
  
        // copy remaining elements of left, if any  
        while (i < len1) { 
            arr[k] = left[i]; 
            k++; 
            i++; 
        } 
  
        // copy remaining element of right, if any  
        while (j < len2) { 
            arr[k] = right[j]; 
            k++; 
            j++; 
        } 
    }
    // iterative Timsort function to sort the  
    // array[0...n-1] (similar to merge sort)  
    public static int[] timSort(int[] arr) {
    	int n = arr.length;
        // Sort individual subarrays of size RUN  
        for (int i = 0; i < n; i += RUN) { 
            insertionSort(arr, i, Math.min((i + 31), (n - 1))); 
        } 
  
        // start merging from size RUN (or 32). It will merge  
        // to form size 64, then 128, 256 and so on ....  
        for (int size = RUN; size < n; size = 2 * size) { 
            
            // pick starting point of left sub array. We  
            // are going to merge arr[left..left+size-1]  
            // and arr[left+size, left+2*size-1]  
            // After every merge, we increase left by 2*size  
            for (int left = 0; left < n; left += 2 * size) { 
                  
                // find ending point of left sub array  
                // mid+1 is starting point of right sub array  
                int mid = Math.min((left + size - 1), (n - 1)); 
                int right = Math.min((left + 2 * size - 1), (n - 1)); 
  
                // merge sub array arr[left.....mid] &  
                // arr[mid+1....right]  
                merge(arr, left, mid, right); 
            } 
        } 
        return arr;
    } 
	
	
	static BufferedOutputStream bos = new BufferedOutputStream(System.out);
	static FastReader scanner = new FastReader();

	static <T> void print(T e) throws IOException {
		bos.write((e+"").getBytes());
	}
	
	static <T> void println(T e) throws IOException {
		bos.write((e+"\n").getBytes());
	}
	static void println() throws IOException {
		bos.write("\n".getBytes());
	}	
	static <T> void print(T[] e) throws IOException {
		print(e[0]);
		String space = " ";
		for(int i = 1; i < e.length; i++) {
			print(space+e[i]);
		}
		println();
	}
	
	public static void main (String[] args) throws IOException{
		 
		 FastReader scanner = new FastReader();
		 final int N = scanner.nextInt();
		 long[] arr = new long[N];
		 long max = Long.MIN_VALUE;
		 for(int i = 0 ; i < N ; i++) {
			 arr[i] = scanner.nextLong();
			 if(arr[i] > max)
				 max = arr[i];
		 }
		 long[] newArr = new long[N];
		 for(int i = 0 ; i < N ; i++) {
			 newArr[i] = max-arr[i];
			 
		 }
		 long gcd = findGCD(newArr);
		
		 long persons = 0;
		 for(int i = 0 ; i < N ; i++) {
			 persons += (max-arr[i])/gcd;
		 }
		 System.out.println(persons+" "+gcd);
		 
	}
	 static long gcd(long a, long b)  { 
       if (a == 0) 
           return b; 
       return gcd(b % a, a); 
    } 
	 
	 static long findGCD(long arr[]) { 
		int n = arr.length;
       long result = arr[0]; 
       for (int i = 1; i < n; i++)  {
       	result = gcd(arr[i], result);
       }
            
 
       return result; 
    } 
	 
}