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;
int n, a[N];
map<int, int> Map;
int calc(int x) {
  if (Map.count(x) == 0) return 0;
  return Map[x];
}
string toStr(unsigned long long a) {
  if (a == 0) return "0";
  string res = "";
  while (a) res = (char)(a % 10 + '0') + res, a /= 10;
  return res;
}
string sub(unsigned long long aa, unsigned long long bb) {
  bool neg = aa < bb;
  if (neg) swap(aa, bb);
  string a = toStr(aa), b = toStr(bb);
  while (a.size() < b.size()) a = '0' + a;
  while (b.size() < a.size()) b = '0' + b;
  int hold = 0;
  for (int i = (int)a.size() - 1; i >= 0; i--) {
    hold = 10 - hold + a[i] - b[i];
    a[i] = (char)(hold % 10 + '0');
    hold = 1 - hold / 10;
  }
  while (a.size() > 1 && a[0] == '0') a.erase(0, 1);
  if (neg) a = '-' + a;
  return a;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  unsigned long long aa = 0, bb = 0;
  for (int i = 1; i <= n; i++) {
    int k1 = calc(a[i]) + calc(a[i] - 1) + calc(a[i] + 1);
    aa += (unsigned long long)(1) * (i - 1 - k1) * a[i];
    Map[a[i]]++;
  }
  Map.clear();
  for (int i = n; i >= 1; i--) {
    int k2 = calc(a[i]) + calc(a[i] - 1) + calc(a[i] + 1);
    bb += (unsigned long long)(1) * (n - i - k2) * a[i];
    Map[a[i]]++;
  }
  cout << sub(aa, bb);
  return 0;
}
