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;
using ll = long double;
const int MAXN = 200001;
ll cumul[MAXN], arr[MAXN];
map<ll, vector<int>> adjList;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long t;
    cin >> t;
    arr[i] = t;
  }
  cumul[0] = arr[0];
  for (int i = 1; i < n; i++) cumul[i] = cumul[i - 1] + arr[i];
  for (int i = 0; i < n; i++) adjList[arr[i]].push_back(i);
  ll sum = 0;
  for (auto& l : adjList) {
    for (auto i : l.second) {
      ll smaller = 0;
      ll larger = 0;
      ll equal = 0;
      if (adjList.find(l.first - 1) != adjList.end()) {
        smaller = adjList[l.first - 1].size() -
                  (lower_bound(adjList[l.first - 1].begin(),
                               adjList[l.first - 1].end(), i) -
                   adjList[l.first - 1].begin());
      }
      if (adjList.find(l.first + 1) != adjList.end()) {
        larger = adjList[l.first + 1].size() -
                 (lower_bound(adjList[l.first + 1].begin(),
                              adjList[l.first + 1].end(), i) -
                  adjList[l.first + 1].begin());
      }
      ll temp = (smaller + larger) * l.first;
      temp += (cumul[n - 1] - ((i - 1 >= 0) ? cumul[i - 1] : 0)) -
              smaller * (l.first - 1) - larger * (l.first + 1) -
              (n - i) * l.first;
      sum += temp;
    }
  }
  cout << setprecision(0) << std::fixed << sum << endl;
  return 0;
}
