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())
arr = [int(x) for x in input().split()]
arr.sort()

x = {}

for i in range(n):
    if arr[i]<=n and arr[i]>=1:
        x[arr[i]] = x.get(arr[i],0)+1
k = 0
moves=0
for i in range(1,n+1):
    #print(arr[k],i,'ss')
    if i not in x.keys():
        y = arr[k]
        if y<=0:
            moves+=abs(y)+i
        else:
            moves+=(y-i)
        k+=1
    elif x[i]>1:
        x[i]-=1
        y = arr[k]
        moves+=(y-i)
        k+=1
    
print(moves)

