Let's denote a function 

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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

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. 
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
set<long double> s;
map<long double, int> mp;
inline long double read() {
  long double x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long double ans, ans2;
long double n, a[N], sz[N], sum[N], cnt;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read(), s.insert(a[i]);
  for (set<long double>::iterator it = s.begin(); it != s.end(); it++)
    mp[*it] = ++cnt;
  for (int i = n; i; i--) {
    long double tmp = 0;
    if (mp.count(a[i] - 1)) tmp += sz[mp[a[i] - 1]];
    if (mp.count(a[i] + 1)) tmp -= sz[mp[a[i] + 1]];
    ans += sum[i + 1] + tmp - (n - i) * a[i];
    ++sz[mp[a[i]]], sum[i] = sum[i + 1] + a[i];
  }
  printf("%.0Lf", ans);
}
