n = int(input())
import sys
inp = list(map(int,sys.stdin.readline().split()))
# 构建回文数组，只需要把原始输入一分为二，然后计算每个相应位置的distance就行了

template = inp[:n//2]
original = [inp[i] for i in range(n-1,n//2-1,-1)]

def dis(s1,s2):
    length = n//2
    distance = []
    for i in range(length):
        distance.append(original[i] - template[i])
    return distance

ori_dis = dis(template,original)


def work():
    length = n//2
    cnt = 0
    # 逐个判断相邻序列的距离，如果同号，那么就取两个序列中距离较小值（同时加1或者减1的操作数量），然后更新这两个序列的距离
    # 同时操作的计数结束后，加上剩下的只能单一操作的所有数，得到ans
    for i in range(length-1):
        if ori_dis[i]*ori_dis[i+1] > 0:
            res = min(abs(ori_dis[i]),abs(ori_dis[i+1]))
            if ori_dis[i] < 0:
                ori_dis[i] += res
                ori_dis[i+1] += res
            else:
                ori_dis[i] -= res
                ori_dis[i+1] -= res
            cnt += res
    for num in ori_dis:
        cnt += abs(num)
    return cnt


print(work())