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 double g = 10.0, eps = 1e-12;
const int N = 600000 + 10, maxn = 200000 + 10, inf = 0x3f3f3f3f;
long long a[N], id[N];
long long num[N];
int n, cnt;
int getid(long long x) {
  int p = lower_bound(id, id + cnt, x) - id;
  return p;
}
int main() {
  cnt = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    id[cnt++] = a[i];
    id[cnt++] = a[i] - 1;
    id[cnt++] = a[i] + 1;
  }
  sort(id, id + cnt);
  cnt = unique(id, id + cnt) - id;
  long long ans[2] = {0, 0}, sum = 0;
  for (long long i = 0; i < n; i++) {
    long long te =
        num[getid(a[i])] + num[getid(a[i] - 1)] + num[getid(a[i] + 1)];
    ans[0] += (i - te) * a[i] - (sum - num[getid(a[i])] * a[i] -
                                 num[getid(a[i] - 1)] * (a[i] - 1) -
                                 num[getid(a[i] + 1)] * (a[i] + 1));
    if (ans[0] >= 1000000000000000000) {
      ans[1]++;
      ans[0] -= 1000000000000000000;
    } else if (ans[0] < 0) {
      ans[1]--;
      ans[0] += 1000000000000000000;
    }
    sum += a[i];
    num[getid(a[i])]++;
  }
  if (ans[1] < 0) {
    ans[1]++;
    ans[0] -= 1000000000000000000;
    if (ans[1] < 0) ans[0] = -ans[0];
  }
  if (ans[1])
    printf("%lld%018lld\n", ans[1], ans[0]);
  else
    printf("%lld\n", ans[0]);
  return 0;
}
