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} 
a,b=map(int,input().split())
aa=[[0]*(b+2)]
for i in range(a):
    it=list(map(int,input().split()))
    aa.append([0]+it+[0])
aa.append([0]*(b+2))
st=False
ans=[]
ss=[[0]*b for i in range(a)]
for i in range(1,a+1):
    for j in range(1,b+1):
     #   print(i,j,"i")
        if aa[i][j]==1 and aa[i][j+1]==1 and aa[i+1][j]==1 and aa[i+1][j+1]==1:

          #  print(i,j)
            
           # print(i,j)
            ss[i-1][j-1]=1
            ss[i-1][j]=1
            ss[i][j]=1
            ss[i][j-1]=1
            ans.append((i,j))

        elif aa[i][j]==1:
            if ss[i-1][j-1]==0:
             #   print(i,j)
                st=True
                break
    if st:
        break
##        if aa[i][j]==1:
##            ar=[[aa[i-1][j],aa[i][j-1],aa[i-1][j-1]],
##                 [aa[i-1][j],aa[i-1][j+1],aa[i][j+1]],
##                 [aa[i][j+1],aa[i+1][j+1],aa[i+1][j]],
##                 [aa[i+1][j],aa[i+1][j-1],aa[i+1][j]]]
##            ar2=[[[i-1,j],[i,j-1],[i-1,j-1]],
##                 [[i-1,j],[i-1,j+1],[i,j+1]],
##                 [[i,j+1],[i+1,j+1],[i+1,j]],
##                 [[i+1,j],[i+1,j-1],[i+1,j]]]
##            if [1,1,1] not in ar:
##                 st=True
##                 break
##           
##          #  print(i,j)
##            ii=ar.index([1,1,1])
##            o=ar2[ii]+[[i,j]]
##           # print(o)
##            o.sort()
##        #    o=o[0]
##          #  print(o,a,b)
##            if o[0][0]>=a or o[0][1]>=b:
##                continue
##            ans.append((o[0][0],o[0][1]))
##    if st:
##        break
if st:
    print(-1)
else:
    ans=set(ans)
    ans=list(ans)
    print(len(ans))
    for i in ans:
        print(*i)
            
