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 maxn = 2e5 + 5;
const long double eps = 1e-22;
pair<int, int> v[maxn];
int n, cnt[maxn];
long double answer, aib[maxn];
inline void update(int val, int pos) {
  for (; pos <= n; pos += (pos & (-pos))) aib[pos] = aib[pos] + val, cnt[pos]++;
}
inline long double countAib(int pos) {
  long double ans = 0;
  for (; pos > 0; pos -= (pos & (-pos))) ans += cnt[pos];
  return ans;
}
inline long double sumAib(int pos) {
  long double ans = 0;
  for (; pos > 0; pos -= (pos & (-pos))) ans += aib[pos];
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int idx;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i].first;
    v[i].second = i;
  }
  sort(v + 1, v + n + 1);
  idx = 1;
  for (int i = 1; i <= n; i++) {
    while (v[idx].first + 1 < v[i].first) {
      update(v[idx].first, v[idx].second);
      idx++;
    }
    answer += v[i].first * countAib(v[i].second) - sumAib(v[i].second);
  }
  memset(aib, 0, sizeof aib);
  memset(cnt, 0, sizeof cnt);
  idx = n;
  for (int i = n; i >= 1; i--) {
    while (v[idx].first - 1 > v[i].first) {
      update(v[idx].first, v[idx].second);
      idx--;
    }
    answer += v[i].first * countAib(v[i].second) - sumAib(v[i].second);
  }
  cout << setprecision(0) << fixed << answer + eps;
  return 0;
}
