954C. Matrix Walk

There is a matrix A of size x × y filled with integers. For every , A_i, 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, ..., a_n 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, ..., a_n (1 ≤ a_i ≤ 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

Copy

8
1 2 3 6 9 8 5 2

Output

YES
3 3

Input

Copy

6
1 2 1 2 5 3

Output

NO

Input

Copy

2
1 10

Output

YES
4 9

Note

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

Also there exist multiple correct answers for both the first and the third examples.

题意：

给一个人旅游路线，求一个二维数组满足所给路线，看看样例1 ，就能明白

思路：

1.相邻的数字只能相差1，或者相差二维数组的列数，所以相邻的数字差最多为2种

2.由于不能停在同一个方格里，所以相邻的数字不能相同

3.类似样例，3*3的二维数组比较特殊，3并不能一步到达4，需要特殊判断

代码：

#include<bits/stdc++.h>
using namespace std;
const int N = 200005;
int arr[N];
int main(){
	int n;
	cin>>n;
	int maxt=1;
	for(int i=0;i<n;i++){
		cin>>arr[i];
		if(i==0) continue;
		if(arr[i]==arr[i-1]){
			//cout<<"1: ";
			cout<<"NO"<<endl;
			return 0;
		}
		maxt=max(maxt,abs(arr[i]-arr[i-1]));
	}
	for(int i=1;i<n;i++){
		int k=abs(arr[i]-arr[i-1]);
		if(k!=1&&k!=maxt){
			//cout<<"2: ";
			cout<<"NO"<<endl;
			return 0;
		}
		if(arr[i]%maxt==0&&arr[i-1]-arr[i]==1&&maxt!=1){
			//cout<<"3: ";
			cout<<"NO"<<endl;
			return 0;
		}
	    else if(arr[i]%maxt==1&&arr[i]-arr[i-1]==1&&maxt!=1){
	    	//cout<<"4: ";
	    	cout<<"NO"<<endl;
	    	return 0;
		}
	}
	cout<<"YES"<<endl;
	cout<<1000000000<<' '<<maxt<<endl;
	return 0;
}