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 preProcess(mat, aux,M,N): 
      
    # Copy first row of mat[][] to aux[][] 
    for i in xrange(0, N, 1): 
        aux[0][i] = mat[0][i] 
  
    # Do column wise sum 
    for i in xrange(1, M, 1): 
        for j in xrange(0, N, 1): 
            aux[i][j] = mat[i][j] + aux[i - 1][j] 
  
    # Do row wise sum 
    for i in xrange(0, M, 1): 
        for j in xrange(1, N, 1): 
            aux[i][j] += aux[i][j - 1] 
  
# A O(1) time function to compute sum of submatrix 
# between (tli, tlj) and (rbi, rbj) using aux[][] 
# which is built by the preprocess function 
def sumQuery(aux, tli, tlj, rbi, rbj): 
      
    # result is now sum of elements  
    # between (0, 0) and (rbi, rbj) 
    res = aux[rbi][rbj] 
  
    # Remove elements between (0, 0) 
    # and (tli-1, rbj) 
    if (tli > 0): 
        res = res - aux[tli - 1][rbj] 
  
    # Remove elements between (0, 0) 
    # and (rbi, tlj-1) 
    if (tlj > 0): 
        res = res - aux[rbi][tlj - 1] 
  
    # Add aux[tli-1][tlj-1] as elements 
    # between (0, 0) and (tli-1, tlj-1) 
    # are subtracted twice 
    if (tli > 0 and tlj > 0): 
        res = res + aux[tli - 1][tlj - 1] 
  
    return res
n,m=map(int,raw_input().split())
aux=[[0 for i in xrange(m)] for j in xrange(n)]

mat=[]
for i in xrange(n):
    mat.append(map(int,raw_input().split()))
preProcess(mat,aux,n,m)
b=[[0 for i in xrange(m)] for j in xrange(n)]
ans=[]
for i in xrange(n-1):
    for j in xrange(m-1):
        if mat[i][j]:
            sm=sumQuery(aux,i,j,i+1,j+1)
            if sm==4:
                b[i][j]=1
                b[i][j+1]=1
                b[i+1][j]=1
                b[i+1][j+1]=1
                ans.append((i,j))
            else:
                if not b[i][j]:
                    print -1
                    exit()
for i in xrange(n):
    if b[i][-1]==0 and mat[i][-1]==1:
        print -1
        exit()
for i in xrange(m):
    if b[-1][i]==0 and mat[-1][i]==1:
        print -1
        exit()
    
print len(ans)
for i in ans:
    print i[0]+1,i[1]+1
