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Colonel has n badges. He wants to give one badge to every of his n soldiers. Each badge has a coolness factor, which shows how much it's owner reached. Coolness factor can be increased by one for the cost of one coin.

For every pair of soldiers one of them should get a badge with strictly higher factor than the second one. Exact values of their factors aren't important, they just need to have distinct factors.

Colonel knows, which soldier is supposed to get which badge initially, but there is a problem. Some of badges may have the same factor of coolness. Help him and calculate how much money has to be paid for making all badges have different factors of coolness.

Input

First line of input consists of one integer n (1 ≤ n ≤ 3000).

Next line consists of n integers a_i (1 ≤ a_i ≤ n), which stand for coolness factor of each badge.

Output

Output single integer — minimum amount of coins the colonel has to pay.

Example

Input

4
1 3 1 4

Output

1

Input

5
1 2 3 2 5

Output

2

Note

In first sample test we can increase factor of first badge by 1.

In second sample test we can increase factors of the second and the third badge by1.

第一种方法:

是从网上看到的,感觉和我的思路一样但是可以通过

#include<iostream>
#include<algorithm>
using namespace std;
const int  MAX = 3000 + 5;
int a[MAX];
int Mi;
int N;

int main()
{
    int temp = -1;
    Mi = 0;
    cin>>N;

    for(int i = 0;i < N;i++)
        cin>>a[i];
    sort(a,a+N);

    for(int i = 0;i < N;i++)
    {
        if(a[i] > temp)
            temp = a[i];
        else
            Mi += ++temp - a[i];
    }
    cout<<Mi<<endl;


    return 0;
}

 

第二种方法:

完全自己的思路,但是不能通过,不知道哪错了

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<iostream>
#include<algorithm>

using namespace std;

const int  MAX = 3000 + 5;
int a[MAX];
int Mi;
int N;

void DP ( int t )
{
    while( a[t] != -1)
    {
        Mi++;
        t++;
        if(t > N ){ cout<<" XXXX "<<endl;break;}
    }
    a[t] = t;
}

int main()
{
    Mi = 0;

    int temp;
    cin>>N;

    memset(a,-1,sizeof(a));

    for(int i = 1;i <= N;i++)
        {
            cin>>temp;
            if( a[temp] == -1 )
                a[temp] = temp;
            else
                DP( temp );
        }
    cout<<Mi<<endl;


    return 0;
}