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>).
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main() {
    int n,sum;
    cin >> n;
    sum=0;
    // cout<<sum<<'\n';
    int a[n];
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        // cout<<a[i]<<'\n';
    }
    sort(a,a+n);
    for(int j=1;j<=n;j++)
    {
        if(a[j-1]<0){
            sum=sum+j-a[j-1];
            // cout<<sum<<'\n';
        }
        else{
            if(a[j-1]>j){
                sum=sum+a[j-1]-j;
                // cout<<sum<<'\n';

            }
            else{
                sum=sum+j-a[j-1];
                // cout<<sum<<'\n';

            }
        }
    }
    cout<<sum;
    
    
	return 0;
}