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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long power(long long a, long long b, long long modi) {
  a %= modi;
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % modi;
    }
    b /= 2;
    a = (a * a) % modi;
  }
  return res;
}
long long a[200000];
long double ans = 0;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    cout << fixed << setprecision(0);
    long long n;
    cin >> n;
    long long sum = 0;
    map<long long, long long> m;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = n - 1; i >= 0; i--) {
      ans += sum - (n - 1 - i) * a[i];
      ans -= m[a[i] + 1];
      ans += m[a[i] - 1];
      m[a[i]]++;
      sum += a[i];
    }
    cout << ans;
  }
  return 0;
}
