Let's denote a function 

<image>

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>
bool debug = 1;
const double PI = acos(-1.0);
const double eps = 1e-9;
using namespace std;
long long v[200100];
long long sum[200100];
map<long long, vector<int> > pos;
long long n;
long long check() {
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      if (abs(v[j] - v[i]) > 1) {
        res += v[j] - v[i];
      }
    }
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    pos[v[i]].push_back(i);
  }
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + v[i];
  }
  long double res = 0;
  for (long long i = 1; i <= n; i++) {
    res += sum[n] - sum[i];
    res -= v[i] * (n - i);
  }
  for (int i = 1; i <= n; i++) {
    long long ant = v[i] - 1;
    long long prox = v[i] + 1;
    if (pos.find(ant) != pos.end()) {
      res += (long double)(pos[ant].end() -
                           lower_bound(pos[ant].begin(), pos[ant].end(), i));
    }
    if (pos.find(prox) != pos.end()) {
      res -= (long double)(pos[prox].end() -
                           lower_bound(pos[prox].begin(), pos[prox].end(), i));
    }
  }
  printf("%.0Lf\n", res);
  return 0;
}
