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 c1;
struct state {
  int h1, a1, h2, a2, level;
  int laststate;
  state() {}
  state(int a, int b, int c, int d, int e, int f)
      : h1(a), a1(b), h2(c), a2(d), level(e), laststate(f) {}
};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int N;
  cin >> N;
  vector<unsigned long long> a(N);
  for (int i = 0; i < N; ++i) {
    cin >> a[i];
  }
  unsigned long long prfx = 0;
  map<int, unsigned long long> removals;
  map<int, unsigned long long> rcount;
  unsigned long long nans = 0, pans = 0;
  for (int i; i < N; ++i) {
    unsigned long long tp = ((unsigned long long)a[i]) * i;
    if (tp > prfx)
      pans += tp - prfx;
    else
      nans += prfx - tp;
    if (removals.count(a[i]) != 0) {
      unsigned long long ot = ((unsigned long long)a[i]) * rcount[a[i]];
      unsigned long long rmls = removals[a[i]];
      if (ot > rmls)
        nans += ot - rmls;
      else
        pans += rmls - ot;
    }
    prfx += a[i];
    if (rcount.count(a[i] - 1) == 0) rcount[a[i] - 1] = 0;
    if (rcount.count(a[i] + 1) == 0) rcount[a[i] + 1] = 0;
    if (rcount.count(a[i]) == 0) rcount[a[i]] = 0;
    if (removals.count(a[i] - 1) == 0) removals[a[i] - 1] = 0;
    if (removals.count(a[i] + 1) == 0) removals[a[i] + 1] = 0;
    if (removals.count(a[i]) == 0) removals[a[i]] = 0;
    removals[a[i] - 1] += a[i];
    removals[a[i] + 1] += a[i];
    removals[a[i]] += a[i];
    ++rcount[a[i] + 1];
    ++rcount[a[i] - 1];
    ++rcount[a[i]];
    if (pans > nans) {
      pans -= nans;
      nans = 0;
    } else {
      nans -= pans;
      pans = 0;
    }
  }
  if (pans > 0 || (pans == 0 && nans == 0))
    cout << pans << endl;
  else
    cout << "-" << nans << endl;
  return 0;
}
