Codeforces 903D Almost Difference

Codeforces 903D Almost Difference

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's denote a function

You are given an array a consisting of n integers. You have to calculate the sum of d(a_i, a_j) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.

Input

The first line contains one integer n (1 ≤ n ≤ 200000) — the number of elements in a.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — elements of the array.

Output

Print one integer — the sum of d(a_i, a_j) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.

Examples

Input

5
1 2 3 1 3

Output

4

Input

4
6 6 5 5

Output

0

Input

4
6 6 4 4

Output

-8

Note

In the first example:

1. d(a₁, a₂) = 0;
2. d(a₁, a₃) = 2;
3. d(a₁, a₄) = 0;
4. d(a₁, a₅) = 2;
5. d(a₂, a₃) = 0;
6. d(a₂, a₄) = 0;
7. d(a₂, a₅) = 0;
8. d(a₃, a₄) =  - 2;
9. d(a₃, a₅) = 0;
10. d(a₄, a₅) = 2.

分析：刚拿到这题，先想到的是：归并排序，仔细构思了一下感觉细节很多，不大好写；转而去想线段树（归并排序能解决的线段树都行），绝对值差在1及以内的直接不考虑就行了，那么对于一个值x，要考虑的区间范围是[1,x-2]U[x+2,+inf]，这样先离散化，然后线段树维护离散化后的下标，维护的信息有两个：区间和与区间计数（sum和cnt），因为对于x的查询结果sum和cnt，ans=cnt*x-sum，然后就是基本的单点更新，区间查询了。这道题要用高精度，我偷懒没写。主要原因是这道题其实用不着线段树，只要set或者map维护一下集合，然后维护前缀和即可。。这样写主程序代码长度不超过50行，比我100行的线段树代码清爽多了。

代码