There is a matrix A of size x × y filled with integers. For every <image>, <image> Ai, j = y(i - 1) + j. Obviously, every integer from [1..xy] occurs exactly once in this matrix. 

You have traversed some path in this matrix. Your path can be described as a sequence of visited cells a1, a2, ..., an denoting that you started in the cell containing the number a1, then moved to the cell with the number a2, and so on.

From the cell located in i-th line and j-th column (we denote this cell as (i, j)) you can move into one of the following cells:

  1. (i + 1, j) — only if i < x; 
  2. (i, j + 1) — only if j < y; 
  3. (i - 1, j) — only if i > 1; 
  4. (i, j - 1) — only if j > 1.



Notice that making a move requires you to go to an adjacent cell. It is not allowed to stay in the same cell. You don't know x and y exactly, but you have to find any possible values for these numbers such that you could start in the cell containing the integer a1, then move to the cell containing a2 (in one step), then move to the cell containing a3 (also in one step) and so on. Can you choose x and y so that they don't contradict with your sequence of moves?

Input

The first line contains one integer number n (1 ≤ n ≤ 200000) — the number of cells you visited on your path (if some cell is visited twice, then it's listed twice).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the integers in the cells on your path.

Output

If all possible values of x and y such that 1 ≤ x, y ≤ 109 contradict with the information about your path, print NO.

Otherwise, print YES in the first line, and in the second line print the values x and y such that your path was possible with such number of lines and columns in the matrix. Remember that they must be positive integers not exceeding 109.

Examples

Input

8
1 2 3 6 9 8 5 2


Output

YES
3 3


Input

6
1 2 1 2 5 3


Output

NO


Input

2
1 10


Output

YES
4 9

Note

The matrix and the path on it in the first test looks like this:

<image>

Also there exist multiple correct answers for both the first and the third examples.
import java.util.*;

import java.io.*;
import java.math.*;
import java.awt.geom.*;

import static java.lang.Math.*;

public class Solution implements Runnable {

	long mod1 = (long) 1e9 + 7;
	int mod2 = 998244353;

	public void solve() throws Exception {
		int n = sc.nextInt();
		int arr[]=sc.readArray(n);
		int ansx=0, ansy=0;
		int currx=0, curry=0;
		int fix=-1;
		for(int i=1;i<n;i++) {
			int diff=Math.abs(arr[i]-arr[i-1]);
			if(Math.abs(diff)==1) continue;
			if(Math.abs(diff)==0) {
				NO();
				return;
			}
			if(fix==-1) {
				fix=diff;
				continue;
			}
			else {
				if(fix!=diff) {
					NO();
					return;
				}
			}
			
		}
		if(fix==-1) {
			out.println("YES");
			out.println((int)1e9+" "+1);
			return;
		}
		for(int i=1;i<n;i++) {
			currx=(arr[i]-1)/fix;
			curry=(arr[i]-1)%fix;
			int prevx=(arr[i-1]-1)/fix;
			int prevy=(arr[i-1]-1)%fix;
			//out.println(prevx+" "+prevy+" "+currx+" "+curry);
			
			if(Math.abs(prevx-currx) + Math.abs(prevy-curry)==1) continue;
			else {
				out.println("NO");
				return;
			}
		}
		
		out.println("YES");
		out.println((int)1e9+" "+fix);

	}
	
	public void NO() {
		out.println("NO");
	}

	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

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

	static long ncr(int n, int r, long p) {
		if (r > n)
			return 0l;
		if (r > n - r)
			r = n - r;

		long C[] = new long[r + 1];

		C[0] = 1;

		for (int i = 1; i <= n; i++) {

			for (int j = Math.min(i, r); j > 0; j--)
				C[j] = (C[j] + C[j - 1]) % p;
		}
		return C[r] % p;
	}

	public long power(long x, long y, long p) {
		long res = 1;
		// out.println(x+" "+y);
		x = x % p;
		if (x == 0)
			return 0;

		while (y > 0) {
			if ((y & 1) == 1)
				res = (res * x) % p;
			y = y >> 1;
			x = (x * x) % p;
		}
		return res;
	}

	static Throwable uncaught;

	BufferedReader in;
	FastScanner sc;
	PrintWriter out;

	@Override
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			sc = new FastScanner(in);
			solve();
		} catch (Throwable uncaught) {
			Solution.uncaught = uncaught;
		} finally {
			out.close();
		}
	}

	public static void main(String[] args) throws Throwable {
		Thread thread = new Thread(null, new Solution(), "", (1 << 26));
		thread.start();
		thread.join();
		if (Solution.uncaught != null) {
			throw Solution.uncaught;
		}
	}

}

class FastScanner {

	BufferedReader in;
	StringTokenizer st;

	public FastScanner(BufferedReader in) {
		this.in = in;
	}

	public String nextToken() throws Exception {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		return st.nextToken();
	}

	public int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}

	public int[] readArray(int n) throws Exception {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long nextLong() throws Exception {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws Exception {
		return Double.parseDouble(nextToken());
	}

}