Permutation *p* is an ordered set of integers *p*<sub class="lower-index">1</sub>,  *p*<sub class="lower-index">2</sub>,  ...,  *p*<sub class="lower-index">*n*</sub>, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p*<sub class="lower-index">*i*</sub>. We'll call number *n* the size or the length of permutation *p*<sub class="lower-index">1</sub>,  *p*<sub class="lower-index">2</sub>,  ...,  *p*<sub class="lower-index">*n*</sub>.

You have a sequence of integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.
The first line contains integer *n* (1 ≤ *n* ≤ 3·10<sup class="upper-index">5</sup>) — the size of the sought permutation. The second line contains *n* integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> ( - 10<sup class="upper-index">9</sup> ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>).
n = int(input())
d = [int(x) for x in input().split()]

a = 0
state = [0 for i in range(n+1)]

for num in d:
    if num < 1:
        a += 1-num
        state[1] += 1
    elif num > n:
        a += num - n
        state[n] += 1
    else:
        state[num] += 1

odd = []
holes = []

for i in range(1, n+1):
    cnti = state[i]
    if cnti == 0:
        holes.append(i)
    if cnti > 1:
        for _ in range(cnti-1):
            odd.append(i)
    while odd and holes:
        fr = odd.pop()
        to = holes.pop()
        a += abs(fr-to)
print(a)