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 INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double E = 2.71828182846;
const int maxn = 200005;
long double ans;
int n;
long double sum[maxn];
int a[maxn];
set<pair<int, int> > s;
vector<pair<int, int> > v;
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    v.push_back(make_pair(a[i], i));
    sum[i] = sum[i - 1] + (long double)a[i];
  }
  sort(v.begin(), v.end());
  for (int i = 1; i < n; i++) {
    int y = a[i];
    long double q1 = 0, q2 = 0, q3 = 0;
    q1 = (lower_bound(v.begin(), v.end(), make_pair(y - 1, n + 1)) - v.begin());
    q1 -=
        (lower_bound(v.begin(), v.end(), make_pair(y - 1, i + 1)) - v.begin());
    q2 = (lower_bound(v.begin(), v.end(), make_pair(y, n + 1)) - v.begin());
    q2 -= (lower_bound(v.begin(), v.end(), make_pair(y, i + 1)) - v.begin());
    q3 = (lower_bound(v.begin(), v.end(), make_pair(y + 1, n + 1)) - v.begin());
    q3 -=
        (lower_bound(v.begin(), v.end(), make_pair(y + 1, i + 1)) - v.begin());
    long double aux = (n - i);
    aux -= (q1 + q2 + q3);
    ans += (sum[n] - sum[i] - (q1 * (y - 1) + q2 * (y) + q3 * (y + 1))) -
           (aux * a[i]);
  }
  printf("%.0Lf\n", ans);
  return 0;
}
