In this problem, a n × m rectangular matrix a is called increasing if, for each row of i, when go from left to right, the values strictly increase (that is, a_{i,1}<a_{i,2}<...<a_{i,m}) and for each column j, when go from top to bottom, the values strictly increase (that is, a_{1,j}<a_{2,j}<...<a_{n,j}).

In a given matrix of non-negative integers, it is necessary to replace each value of 0 with some positive integer so that the resulting matrix is increasing and the sum of its elements is maximum, or find that it is impossible.

It is guaranteed that in a given value matrix all values of 0 are contained only in internal cells (that is, not in the first or last row and not in the first or last column).

Input

The first line contains integers n and m (3 ≤ n, m ≤ 500) — the number of rows and columns in the given matrix a.

The following lines contain m each of non-negative integers — the values in the corresponding row of the given matrix: a_{i,1}, a_{i,2}, ..., a_{i,m} (0 ≤ a_{i,j} ≤ 8000).

It is guaranteed that for all a_{i,j}=0, 1 < i < n and 1 < j < m are true.

Output

If it is possible to replace all zeros with positive numbers so that the matrix is increasing, print the maximum possible sum of matrix elements. Otherwise, print -1.

Examples

Input


4 5
1 3 5 6 7
3 0 7 0 9
5 0 0 0 10
8 9 10 11 12


Output


144


Input


3 3
1 2 3
2 0 4
4 5 6


Output


30


Input


3 3
1 2 3
3 0 4
4 5 6


Output


-1


Input


3 3
1 2 3
2 3 4
3 4 2


Output


-1

Note

In the first example, the resulting matrix is as follows: 
    
    
      
    1 3 5 6 7  
    3 6 7 8 9  
    5 7 8 9 10  
    8 9 10 11 12  
    

In the second example, the value 3 must be put in the middle cell.

In the third example, the desired resultant matrix does not exist.
import java.util.Scanner;  // Import the Scanner class

public final class Main{
    static int[] tempList;
    public static void main(String[ ] args){    
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt(), n = sc.nextInt();
        int[][] tab = new int[m][n];
        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++){
                tab[i][j] = sc.nextInt();
            }
        }

        int sum=0;


        for(int i=m-1;i>=0;i--) for(int j=n-1;j>=0;j--){
            if(i<m-1 && j<n-1 &&tab[i][j]==0)
                tab[i][j] = min(tab[i+1][j],tab[i][j+1])-1;
            else if((i<m-1 && tab[i][j]>=tab[i+1][j] )|| (j<n-1 && tab[i][j]>=tab[i][j+1])){
                System.out.println(-1);
                return;
            }

            sum+=tab[i][j];
                
        }

        System.out.println(sum);
        sc.close();;

        /*for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++){
                System.out.print(tab[i][j] + " ");
            }
            System.out.print("\n");
        }*/
        
    }

    
    static int min(int x, int y){
        return x>y ? y : x;
    }
    
    
}