import math

N = int(input())
grid = []
for _ in range(8):
    l = list( map(int, input().split()) )
    grid.append(l)

pre_sum = [[val for val in grid[i]] for i in range(8)]
for i in range(8):
    for j in range(1, 8):
        pre_sum[i][j] += pre_sum[i][j-1]
for i in range(1, 8):
    for j in range(8):
        pre_sum[i][j] += pre_sum[i-1][j]

def get_sum(i1, j1, i2, j2):
    val1 = 0 if i1-1 < 0 else pre_sum[i1-1][j2]
    val2 = 0 if j1-1 < 0 else pre_sum[i2][j1-1]
    val3 = 0 if i1-1 < 0 or j1-1 < 0 else pre_sum[i1-1][j1-1]
    return pre_sum[i2][j2] - val1 - val2 + val3

# 经过公式化简，目标是要让每个分块的xi的平方和最小
# dp(i1, j1, i2, j2)表示左上角是i1, j1 右下角是i2, j2的矩形切分k次的所有方案的分块价值平方和的最小值

from functools import lru_cache
@lru_cache(typed=False, maxsize=128000000)
def dp(i1, j1, i2, j2, k):

    if k == 0:
        return get_sum(i1, j1, i2, j2)*get_sum(i1, j1, i2, j2)

    if i2 == i1 and j2 == j1:
        # 无法切分
        return 0x7fffffff

    # 纵着切分
    ans = 0x7fffffff
    for i in range(i1, i2):
        val1 = get_sum(i1, j1, i, j2)*get_sum(i1, j1, i, j2) + dp(i+1, j1, i2, j2, k-1)
        val2 = get_sum(i+1, j1, i2, j2)*get_sum(i+1, j1, i2, j2) + dp(i1, j1, i, j2, k-1)
        ans = min(ans, val1, val2)

    # 横着切分
    for j in range(j1, j2):
        val1 = get_sum(i1, j1, i2, j)*get_sum(i1, j1, i2, j) + dp(i1, j+1, i2, j2, k-1)
        val2 = get_sum(i1, j+1, i2, j2)*get_sum(i1, j+1, i2, j2) + dp(i1, j1, i2, j, k-1)
        ans = min(ans, val1, val2)

    return ans

total = 0
for i in range(8):
    total += sum(grid[i])
mean = total / N

square_sum = dp(0, 0, 7, 7, N-1)
ans = math.sqrt((square_sum - N * mean * mean) / N)

print('%.3f' % ans)