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("omit-frame-pointer")
#pragma GCC optimize("unroll-loops")
using namespace std;
string itosm(long long x) {
  if (x == 0) return "0";
  string ans = "";
  while (x > 0) {
    ans += ((x % 10) + '0');
    x /= 10;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
long long stoim(string str) {
  long long ans = 0;
  long long k = 1;
  for (int i = str.length() - 1; i >= 0; i--) {
    ans += (str[i] - '0') * k;
    k *= 10;
  }
  return ans;
}
const long long infll = 1e18 + 3;
const int inf = 1009000999;
const double eps = 1e-7;
const int maxn = 1e6 + 46;
const int baseint = 1000200013;
const long long basell = 1e18 + 3;
const long double PI = acos(-1.0);
const long long mod = 1e9 + 7;
long long a[maxn];
map<long long, long long> cnt;
void solve() {
  int n;
  cin >> n;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
    cnt[a[i]]++;
  }
  long double ans = 0, fcnt, fsum;
  for (int i = 0; i < n; i++) {
    fcnt = n - i - 1;
    sum -= a[i];
    cnt[a[i]]--;
    fsum = sum;
    fcnt -= (cnt[a[i]] + cnt[a[i] - 1] + cnt[a[i] + 1]);
    fsum -= (cnt[a[i]] * a[i] + cnt[a[i] - 1] * (a[i] - 1) +
             cnt[a[i] + 1] * (a[i] + 1));
    ans += (fsum - a[i] * fcnt);
  }
  cout << fixed << setprecision(0) << ans;
}
int main() {
  srand(time(0));
  ios_base::sync_with_stdio(0);
  ;
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
