//条件为一个数+1一个数-1，即总和不变，先判断总和再排序后依次取各个数差的绝对值除2 #include<iostream>

#include<cmath>

#include<algorithm>

using namespace std;

#define ll long long

int main()

{

ios::sync_with_stdio(0);

cin.tie(0);

cout.tie(0);

ll n,sum1=0;

cin>>n;

ll a[n+10];

for(int i=1;i<=n;i++){

cin>>a[i];

sum1+=a[i];

}

ll sum2=(n+n*n)/2;

if(sum1!=sum2){

cout<<-1<<"\n";

}else{

ll ans=0;

sort(a+1,a+n+1);

for(int i=1;i<=n;i++){

if(a[i]!=i){

ans=ans+abs(a[i]-i);

}

}

cout<<ans/2<<"\n";

}

return 0;

}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e6+5;
ll a[N];
void solve(){
	ll n;cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	sort(a+1,a+1+n);
	ll sf=0,sz=0;
	for(int i=1;i<=n;i++){
		if(a[i]<i) sf+=i-a[i];
		else sz+=a[i]-i;
	}
	if(sz!=sf) cout<<"-1"<<'\n';
	else cout<<sf<<'\n';
}
int main(){
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	ll _=1;
//	cin>>_;
	while(_--){
		solve();
	}
	return 0;
}