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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeMap;
import java.util.TreeSet;

import sun.security.jgss.TokenTracker;
 
public class MyCode
{
	public static void main(String args[])
	{
	   Scanner sc = new Scanner(System.in); 
	   int i,j,n,m,x,element;
	   Integer[][] arr = new Integer[600][600];
	   Integer[][] zero = new Integer[600][600];
	   while(sc.hasNextInt())
	   {
		   n = sc.nextInt();
		   m = sc.nextInt();
		   for(i=0;i<n;i++)
		   {
			   for(j=0;j<m;j++)
			   {
				  zero[i][j]=0;
				  x= sc.nextInt();
				  arr[i][j]=x;
				  if(x==0)
					  zero[i][j]=1;
			   }
		   }
		
		   for(j=m-2;j>0;j--)
		   {
			   for(i=n-2;i>=1;i--)
			   {
				   if(arr[i][j]==0)
				   {
					   arr[i][j]=arr[i+1][j]-1;
				   }
				  
			   }
		   }
		  
		   for(i=1;i<n-1;i++)
		   {
			   for(j=m-2;j>0;j--)
			   {
				  if(zero[i][j]==1)
				  {
					  element = arr[i][j];
					  if(arr[i][j-1]<element && element<arr[i][j+1])continue;
					  else
					  {
						  element = arr[i][j+1]-1;
						  arr[i][j]= Math.min(arr[i][j], element);
					  }
				  }
			   }
		   }
		   boolean ans=true;
		   for(i=0;i<n;i++)
		   {
			   for(j=0;j<m;j++)
			   {
				   if(j>0 && arr[i][j]<=arr[i][j-1])
				   {
					   ans=false;
					   break;
				   }
				   if(i>0 && arr[i][j]<=arr[i-1][j])
				   {
					   ans=false;
					   break;
				   }
			   }
		   }
		   
		   
		   if(ans==false)
		   {
			   System.out.println("-1");
			   continue;
		   }
		   long total=0;
		   for(i=0;i<n;i++)
		   {
			   for(j=0;j<m;j++)
			   {
				  total+=arr[i][j];
			   }
		   }
		   if(ans)
			   System.out.println(total);
	   }
	   
	}
	
	   
}

