codeforces 903D

Let's denote a function

$\text{[math]}$

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

5
1 2 3 1 3

output

input

4
6 6 5 5

output

input

4
6 6 4 4

output

-8

Note

In the first example:

d(a1, a2) = 0;
d(a1, a3) = 2;
d(a1, a4) = 0;
d(a1, a5) = 2;
d(a2, a3) = 0;
d(a2, a4) = 0;
d(a2, a5) = 0;
d(a3, a4) = - 2;
d(a3, a5) = 0;
d(a4, a5) = 2.

题意很好理解，这里就不说了。

解题思路：由给出的公式可以得到d(a1,an)+d(a2,an)+d(a3,an)+...+d(ai,an)的答案是i*an-sum (sum是a1~ai中与an差的绝对值大于1的数之和，那么这样只需要遍历一遍数组，执行i*an-sum并减去a1~ai中与an差的绝对值小于等于1的数就行了。所以还要用map记录一下。

坑点：注意数据1e9,1e9-2,1e9-4...-1e9+4,-1e9+2,-1e9 (共有200000个，这种时候答案近似于100000*100000*1e9 数据会爆long long,所以要用long double

附ac代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <string>
 4 #include <iostream>
 5 #include <algorithm>
 6 #include <map>
 7 #include <iomanip>
 8 using namespace std;
 9 typedef long double ld;
10 const int maxn = 200000+10;
11 ld nu[maxn];
12 map<ld,ld>mp;
13 int main()
14 {
15     int n;
16     scanf("%d",&n);
17     for(int i=0;i<n;++i)
18         cin>>nu[i];
19     ld s=0;
20     ld sum=0;
21     mp.clear();
22     for(int i=0;i<n;++i) //根据推出的公式可得，i从0开始比较简单
23     {
24         s+=i*nu[i]-sum+mp[nu[i]+1]-mp[nu[i]-1];
25         //s+=((i-mp[nu[i]-1]-mp[nu[i]+1]-mp[nu[i]])*nu[i]-(sum-mp[nu[i]+1]*(nu[i]+1)-mp[nu[i]-1]*(nu[i]-1)-mp[nu[i]]*[i]));
26         mp[nu[i]]++;
27         sum+=nu[i];
28     }
29     cout<<fixed<<setprecision(0)<<s<<endl;
30     return 0;
31 }

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