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 = 200010;
int a[N], n, b[N];
long double pre[N] = {0}, ans = 0;
vector<int> ve[N];
void solve(int t, int p, int v) {
  int l = lower_bound(ve[t].begin(), ve[t].end(), p) - ve[t].begin();
  ans += v * l;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pre[i] = pre[i - 1] + a[i];
    ans += (i - 1) * (long double)a[i] - pre[i - 1];
    b[i - 1] = a[i];
  }
  sort(b, b + n);
  int m = unique(b, b + n) - b;
  unordered_map<int, int> mp;
  for (int i = 0; i < m; ++i) mp[b[i]] = i;
  for (int i = 1; i <= n; ++i) {
    int t = mp[a[i]];
    ve[t].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    int t = a[i];
    if (mp.count(t + 1)) solve(mp[t + 1], i, 1);
    if (mp.count(t - 1)) solve(mp[t - 1], i, -1);
  }
  cout << fixed << setprecision(0) << ans << endl;
  return 0;
}
