You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.

You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 × 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 ≤ x < n and 1 ≤ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.

Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.

Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 50).

Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.

Output

If it is impossible to make B equal to A, print one integer -1.

Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k — the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 ≤ k ≤ 2500 should hold.

Examples

Input


3 3
1 1 1
1 1 1
0 1 1


Output


3
1 1
1 2
2 2


Input


3 3
1 0 1
1 0 1
0 0 0


Output


-1


Input


3 2
0 0
0 0
0 0


Output


0

Note

The sequence of operations in the first example:

\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & → & 1 & 1 & 0 & → & 1 & 1 & 1 & → & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} 


def solve(A):

    nrSqMatches = 0
    nrOnes = 0
    coords = []
    for i in range(len(A)):
        for j in range(len(A[i])):
            if i < len(A)-1 and j < len(A[i])-1:
                tmp = A[i][j] + A[i+1][j] + A[i][j+1] + A[i+1][j+1]
                if tmp == 4:
                    nrSqMatches += 1
                    coords.append([i+1,j+1])
            if A[i][j] == 1:
                nrOnes += 1

            if not isOK(A,i,j):
                return [[-1]]

    if nrSqMatches*4 < nrOnes:
        return [[-1]]



    return coords

def isOK(A,i,j):

    left = False
    right = False
    up = False
    down = False
    center = A[i][j]

    if i > 0:
        left = A[i-1][j]==1
    if j > 0:
        up = A[i][j-1]==1
    if i < len(A)-1:
        right = A[i+1][j]==1
    if j <len(A[i])-1:
        down = A[i][j+1]==1
    
    if center == 0:
        return True
    
    if up and (left or right):
        return True
    
    if down and (left or right):
        return True

    return False

if __name__=='__main__':
    (n,m) = map(lambda x:int(x), input().strip().split())
    A = []
    for i in range(n):
        row = list(map(lambda x:int(x), input().strip().split()))
        A.append(row)

    #print(A)
    sol = solve(A)

    if len(sol)==0:
        print(0)
    elif sol[0][0]==-1:
        print(-1)
    else:
        print(len(sol))
        for row in sol:
            line = ''
            for v in row:
                line += str(v) + ' '
            print(line)



