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;
typedef  long long  ll;
typedef vector<ll> vll;
#define rall(s) s.rbegin(),s.rend()
typedef map<ll,ll> mll;
#define fr(n,x) for(ll n=0;n<x;n++)
#define pb push_back
#define all(s) s.begin(),s.end()
#define EL cout<<endl
const ll MOD=1e9+7;


int main(){
    ll x,n,k,d;
    cin>>x;
    ll res=0;
    vll s;
    fr(n,x){
        cin>>d;
        if(d<=0){
            res+=(1-d);
            s.pb(1);
        }
        else if(d>x){
            res+=(d-x);
            s.pb(x);
        }
        else s.pb(d);
    }
    sort(all(s));
    fr(n,x-1){
        if(s[n]==s[n+1]){
            res++;
            s[n+1]=s[n]+1;
        }
    }
    cout<<res;
}