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;
long long n;
long double a[200100], b[200100];
long double t[500100 * 4], sz, t1[500100 * 4];
map<int, int> mp1;
void update(int v, long double x) {
  v = v + sz - 1;
  t[v] += x;
  t1[v]++;
  v >>= 1;
  while (v) {
    t[v] = t[v + v] + t[v + v + 1];
    t1[v] = t1[v + v] + t1[v + v + 1];
    v >>= 1;
  }
}
long double get(int v, int l, int r, int tl, int tr) {
  if (l > r || tl > r || l > tr || tl > tr) return 0;
  if (tl <= l && r <= tr) {
    return t[v];
  }
  int mid = (l + r) >> 1;
  return get(v + v, l, mid, tl, tr) + get(v + v + 1, mid + 1, r, tl, tr);
}
long double get1(int v, int l, int r, int tl, int tr) {
  if (l > r || tl > r || l > tr || tl > tr) return 0;
  if (tl <= l && r <= tr) {
    return t1[v];
  }
  int mid = (l + r) >> 1;
  return get1(v + v, l, mid, tl, tr) + get1(v + v + 1, mid + 1, r, tl, tr);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mp1[a[i]] = 1;
    mp1[a[i] - 2] = 1;
    mp1[a[i] + 2] = 1;
  }
  int cur = 0;
  for (auto it = mp1.begin(); it != mp1.end(); it++) {
    cur++;
    it->second = cur;
  }
  sz = 1;
  while (sz < cur) sz += sz;
  for (int i = 1; i <= n; i++) {
    b[i] = mp1[a[i]];
  }
  long double ans = 0;
  for (int i = 1; i <= n; i++) {
    long double sum =
        get(1, 1, sz, 1, mp1[a[i] - 2]) + get(1, 1, sz, mp1[a[i] + 2], cur);
    long double kol =
        get1(1, 1, sz, 1, mp1[a[i] - 2]) + get1(1, 1, sz, mp1[a[i] + 2], cur);
    ans += (a[i] * kol - sum);
    update(b[i], a[i]);
  }
  cout.precision(0);
  cout << fixed << ans;
}
