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;
int ini() {
  int x;
  scanf("%d", &x);
  return x;
}
long long inl() {
  long long x;
  scanf("%lld", &x);
  return x;
}
long long arr[200005], suf[200005];
map<long long, long long> mp1;
int main() {
  int n = ini();
  for (int i = 1; i <= n; i++) arr[i] = inl();
  long double res = 0;
  for (int i = n; i >= 1; i--) {
    suf[i] = suf[i + 1] + arr[i];
    long long cn = n - i;
    long long ans = 0;
    if (mp1.find(arr[i] - 1) != mp1.end()) {
      cn -= mp1[arr[i] - 1];
      ans -= mp1[arr[i] - 1] * (arr[i] - 1);
    }
    if (mp1.find(arr[i]) != mp1.end()) {
      cn -= mp1[arr[i]];
      ans -= mp1[arr[i]] * (arr[i]);
    }
    if (mp1.find(arr[i] + 1) != mp1.end()) {
      cn -= mp1[arr[i] + 1];
      ans -= mp1[arr[i] + 1] * (arr[i] + 1);
    }
    ans += suf[i + 1];
    ans -= (cn * arr[i]);
    res += ans;
    mp1[arr[i]]++;
  }
  printf("%0.0Lf\n", res);
  return 0;
}
