Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an *n* × *n* matrix *W*, consisting of integers, and you should find two *n* × *n* matrices *A* and *B*, all the following conditions must hold: 
 -  *A*<sub class="lower-index">*ij*</sub> = *A*<sub class="lower-index">*ji*</sub>, for all *i*, *j* (1 ≤ *i*, *j* ≤ *n*); -  *B*<sub class="lower-index">*ij*</sub> =  - *B*<sub class="lower-index">*ji*</sub>, for all *i*, *j* (1 ≤ *i*, *j* ≤ *n*); -  *W*<sub class="lower-index">*ij*</sub> = *A*<sub class="lower-index">*ij*</sub> + *B*<sub class="lower-index">*ij*</sub>, for all *i*, *j* (1 ≤ *i*, *j* ≤ *n*). 
Can you solve the problem?
The first line contains an integer *n* (1 ≤ *n* ≤ 170). Each of the following *n* lines contains *n* integers. The *j*-th integer in the *i*-th line is *W*<sub class="lower-index">*ij*</sub> (0 ≤ |*W*<sub class="lower-index">*ij*</sub>| &lt; 1717).
# input
n = int(input())
matrix = []
a = n
while a>0:
    x = []
    x = input().split()
    matrix.append(x)
    a -= 1

# A
a = 0
while a<n:
    b = 0
    out = ''
    while b<n:
        if a == b:
            if b == 0:
                out = str(float(matrix[a][b]))
            else:
                out += ' ' + str(float(matrix[a][b]))
        else:
            if b == 0:
                out = str((float(matrix[a][b]) + float(matrix[b][a]))/2)
            else:
                out += ' ' + str((float(matrix[a][b]) + float(matrix[b][a]))/2)
        b += 1
    print(out)
    a += 1


# B
a = 0
while a<n:
    b = 0
    out = ''
    while b<n:
        if a == b:
            if b == 0:
                out = str(float(0))
            else:
                out += ' ' + str(float(0))
        else:
            if b == 0:
                out = str((float(matrix[a][b]) - float(matrix[b][a]))/2)
            else:
                out += ' ' + str((float(matrix[a][b]) - float(matrix[b][a]))/2)
        b += 1
    print(out)
    a += 1