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;
void display(vector<int> v1) {
  for (int i = 0; i < v1.size(); i++) {
    cout << v1[i] << " ";
  }
  cout << endl;
}
long long int mod_expo(long long int MOD, long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b % 2) {
      ans *= a;
      ans %= MOD;
    }
    b /= 2;
    a *= a;
    a %= MOD;
  }
  return ans % MOD;
}
int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[8] = {1, 0, -1, 0, 1, -1, 1, -1};
using namespace std;
const int MAXN = (int)2e5 + 1000;
long double suff[MAXN];
long double arr[MAXN];
int n;
map<long double, long double> m1;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i += 1) {
    cin >> arr[i];
    m1[arr[i]] += 1;
    suff[i] = arr[i];
  }
  for (int i = n; i >= 1; i -= 1) {
    suff[i] += suff[i + 1];
  }
  long double ans = 0;
  for (int i = 1; i <= n; i += 1) {
    long double same = m1[arr[i]];
    long double oneabove = m1[arr[i] + 1];
    long double onebelow = m1[arr[i] - 1];
    long double nosremain = n - i + 1;
    nosremain -= same;
    nosremain -= oneabove;
    nosremain -= onebelow;
    long double summa = suff[i];
    summa -= (same * arr[i]);
    summa -= (oneabove * (arr[i] + 1));
    summa -= (onebelow * (arr[i] - 1));
    ans += (summa - (nosremain * arr[i]));
    m1[arr[i]] -= 1;
  }
  cout << setprecision(0);
  cout << fixed << ans << endl;
  return 0;
}
