Codeforces Round #304 (Div. 2) B. Soldier and Badges【思维/给你一个序列，每次操作你可以对一个元素加1，问最少经过多少次操作，才能使所有元素互不相同】

B. Soldier and Badges

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Colonel has n badges. He wants to give one badge to every of his n soldiers. Each badge has acoolness 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 ai (1 ≤ ai ≤ n), which stand for coolness factor of each badge.

Output

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

Examples

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 by 1.

【题意】：给你一个序列，每次操作你可以对一个元素加1，问最少经过多少次操作，才能使所有元素互不相同

【分析】：排序，求和。有相邻相等的，后面的+1。后面的小于前面的，后面的变为前面的数+1，再次求新和。新和-旧和为所求。

【代码】：

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
int sum1,sum2;
int a[3000];
int main()
{
    int n, i, j;

    scanf("%d", &n);
    for (i = 0; i < n; i++)
    {
        scanf("%d", &a[i]);
        sum1+=a[i];
    }

    sort(a, a + n);
    sum2 = a[0];
    for (i = 1; i < n; i++)
    {
        if(a[i]==a[i-1]) 
            a[i]++;
            
        else if(a[i]<a[i-1]) 
            a[i] = a[i-1]+1;
        
        sum2+=a[i];
    }
    printf("%d
", sum2-sum1);
    return 0;
}
/*
1 2 2 3 5=13
1 2 3 4 5=15
15-13=2
1 1 1 2 5
1 2 3 4 5

*/