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.
def ii():
  return int(input())
def ss():
  return [x for x in input()]
def si():
  return [int(x) for x in input().split()]
def mi():
  return map(int, input().split())

a, b = mi()
s = []
ans = 0
key = True

for i in range(a):
	i = si()
	s.append(i)
	ans += sum(i)

for i in range(a - 1):
	if not (s[i][0] < s[i + 1][0] and s[i][-1] < s[i + 1][-1]):
		ans = -1
		key = False
		break
for i in range(b - 1):
	if not (s[0][i] < s[0][i + 1] and s[-1][i] < s[-1][i + 1]):
		ans = -1
		key = False
		break

for i in range(a - 2, 0, -1):
	if not key:
		break
	for x in range(b - 2, 0, -1):
		if s[i][x] < s[i + 1][x] and s[i][x] < s[i][x + 1] and (min(s[i + 1][x], s[i][x + 1]) - max(s[i - 1][x], s[i][x - 1])) > 1:
			if s[i][x] == 0:
				s[i][x] = min(s[i + 1][x], s[i][x + 1]) - 1
				ans += s[i][x]
		else:
			ans = -1
			key = False
			break
print(ans)