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;
struct lll {
  long long a1, a2;
  lll(long long b1 = 0, long long b2 = 0) {
    a1 = b1;
    a2 = b2;
  }
  lll operator+(long long x) const {
    long long b1 = a1;
    long long b2 = a2 + x;
    b1 -= (long long)1e6;
    b2 += (long long)1e18;
    b1 += b2 / (long long)1e12;
    b2 %= (long long)1e12;
    return lll(b1, b2);
  }
  void show() {
    if (a1 == 0) {
      cout << a2 << endl;
    } else if (a1 > 0) {
      cout << a1 << setfill('0') << setw(12) << a2 << endl;
    } else {
      lll t(-a1, 0);
      t = t + (-a2);
      cout << "-";
      t.show();
    }
  }
};
int n;
long long x;
lll ans;
map<long long, long long> m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    long long tmp = x * (2 * i - n - 1) - m[x - 1] + m[x + 1];
    m[x]++;
    ans = ans + tmp;
  }
  ans.show();
  return 0;
}
