Educational Codeforces Round 34 (Rated for Div. 2) D

D. 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(ai, aj) 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, ..., an (1 ≤ ai ≤ 109) — elements of the array.

Output

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

Examples

input

51 2 3 1 3

output

4

input

46 6 5 5

output

0

input

46 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.

此题爆了long long（9*1e18）,可以考虑用long double存储，输出用%Lf，但需要在C++14下才可以

考虑每个数对答案的贡献，即被当做x和y的次数，然后再处理一下特殊的情况即可

蠢哭了

看了一天

#include<bits/stdc++.h>using namespace std;#define maxn 200000+1000#define LL long longlong long a[maxn];map<long long,int>q;int main(){   int n;   cin>>n;   long double sum=0;   for(int j=1;j<=n;j++){      cin>>a[j];      sum+=(j-1)*a[j];      sum-=(n-j)*a[j];   }   for(int j=1;j<=n;j++){      q[a[j]]++;      sum-=q[a[j]-1];      sum+=q[a[j]+1];   }   printf("%.0Lf\n",sum);}//-9999999990000000000   longlong  爆这组