codeforces285C

Building Permutation

 CodeForces - 285C 

Permutation p is an ordered set of integers p₁,  p₂,  ...,  p_n, 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_i. We'll call number n the size or the length of permutation p₁,  p₂,  ...,  p_n.

You have a sequence of integers a₁, a₂, ..., a_n. 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.

Input

The first line contains integer n (1 ≤ n ≤ 3·10⁵) — the size of the sought permutation. The second line contains n integers a₁, a₂, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

Print a single number — the minimum number of moves.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

2
3 0

Output

2

Input

3
-1 -1 2

Output

6

Note

In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).

In the second sample you need 6 moves to build permutation (1, 3, 2).

sol：较显然的是最小的变成1，次小的变成2，以此类推，应该是最优的

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=300005;
int n;
ll a[N];
int main()
{
    int i;
    ll ans=0;
    R(n);
    for(i=1;i<=n;i++) R(a[i]);
    sort(a+1,a+n+1);
    for(i=1;i<=n;i++) ans+=abs(i-a[i]);
    Wl(ans);
    return 0;
}
/*
Input
2
3 0
Output
2

Input
3
-1 -1 2
Output
6
*/

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