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 N = 2e5 + 69;
int n;
__int128 ans;
map<int, int> mp;
int ar[N];
string to_string(__int128 x) {
  if (x == 0) return "0";
  string ret = "";
  bool neg = false;
  if (x < 0) {
    neg = true;
    x = -x;
  }
  while (x > 0) {
    ret += (char)((x % 10) + 48);
    x /= 10;
  }
  if (neg) ret += '-';
  reverse(ret.begin(), ret.end());
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (__int128 i = 1; i <= n; i++) {
    cin >> ar[i];
    ans += 1LL * (2 * i - 1 - n) * ar[i];
  }
  for (int i = n; i >= 1; i--) {
    ans -= mp[ar[i] + 1];
    ans += mp[ar[i] - 1];
    mp[ar[i]]++;
  }
  cout << to_string(ans) << "\n";
  return 0;
}
