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 javafx.util.Pair;

import javax.swing.text.MaskFormatter;
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main {



    public static void main(String[] args){

        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        long A[]=new long[n];
        long max=Integer.MIN_VALUE;
        for(int i=0;i<n;i++)
        {
            A[i]=in.nextInt();
            max=Math.max(A[i],max);
        }
        long B[]=new long[n];
        for(int i=0;i<n;i++)
        {
            B[i]=max-A[i];
        }
        long sum=0;
        long gcd=findGCD(B,n);
        for(int i=0;i<n;i++)
        {
            sum=sum+(max-A[i])/gcd;
        }
        System.out.println(sum+" "+gcd);

    }

    static boolean FindPoint(int x1, int y1, int x2,
                             int y2, int x, int y)
    {
        if (x > x1 && x < x2 &&
                y > y1 && y < y2)
            return true;

        return false;
    }

    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    // Function to find gcd of array of
    // numbers
    static long findGCD(long arr[], int n)
    {
        long result = arr[0];
        for (int i = 1; i < n; i++)
            result = gcd(arr[i], result);

        return result;
    }






/*
for (int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {

            }
        }
 */







    static long largestPower(int n, long p) {
        // Initialize result
        long x = 0;

        // Calculate x = n/p + n/(p^2) + n/(p^3) + ....
        while (n > 0) {
            n /= p;
            x += n;
        }
        return x;
    }

    // Utility function to do modular exponentiation.
// It returns (x^y) % p
    static long power(long x, long y, long p) {
        long res = 1; // Initialize result
        x = x % p; // Update x if it is more than or
        // equal to p
        while (y > 0) {
            // If y is odd, multiply x with result
            if (y % 2 == 1) {
                res = (res * x) % p;
            }

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }
        return res;
    }

    // Returns n! % p
    static long modFact(int n, long p) {
        if (n >= p) {
            return 0;
        }

        long res = 1;

        // Use Sieve of Eratosthenes to find all primes
        // smaller than n
        boolean isPrime[] = new boolean[n + 1];
        Arrays.fill(isPrime, true);
        for (int i = 2; i * i <= n; i++) {
            if (isPrime[i]) {
                for (int j = 2 * i; j <= n; j += i) {
                    isPrime[j] = false;
                }
            }
        }

        // Consider all primes found by Sieve
        for (int i = 2; i <= n; i++) {
            if (isPrime[i]) {
                // Find the largest power of prime 'i' that divides n
                long k = largestPower(n, i);

                // Multiply result with (i^k) % p
                res = (res * power(i, k, p)) % p;
            }
        }
        return res;
    }
    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[64]; // line length
            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();
        }


    }

    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());
        }

    }
}