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())

a=[int(x) for x in input().split()]
# print(a)
countPos=0
neg=[]
pos=[]

countDif=0
for i in range(0, len(a)):
    if a[i]>n:
        countDif+=a[i]-n
        a[i]=n
for i in a:
    if i<=0:
        neg.append(i)
    else:
        pos.append(i)
i=0
pos.sort()

while i < len(pos)-1:

    if pos[i]==pos[i+1] :
        countPos+=pos[i]
        pos.pop(i)
    else:
        i+=1

# print(int(n*(n+1)/2)-sum(pos))
# print(countDif,sum(neg),int(n*(n+1)/2)-sum(pos),countPos)
# print(countPos-(int(n*(n+1)/2)-sum(pos))-sum(neg)) 
print(int(n*(n+1)/2)-sum(pos)+countDif-sum(neg))

# 3 zu 2 =1
# 3 zu 0 = 3 , 0 zu 2 = 2 => 3-2=1 
# 3 zu 1 =2
# 3 zu 0 = 3 , 0 zu 2 = 2 => 3-1=2 
