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 long long maxn = 6e5 + 100;
const long long mx = 1e9 + 10;
long long n, a[maxn];
long double sum, ans, cc[maxn], tree[maxn];
vector<long double> v;
long long getid(long long x) {
  return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
};
long long lowbit(long long x) { return x & (-x); }
void add(long long id, long double x) {
  for (long long i = id; i <= maxn; i += lowbit(i)) tree[i] += x;
}
long double getsum(long long x) {
  long double ret = 0;
  for (long long i = x; i; i -= lowbit(i)) ret += tree[i];
  return ret;
}
void adds(long long id) {
  for (long long i = id; i <= maxn; i += lowbit(i)) cc[i]++;
}
long double getsums(long long x) {
  long double ret = 0;
  for (long long i = x; i; i -= lowbit(i)) ret += cc[i];
  return ret;
}
int main() {
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    v.push_back(a[i]);
    v.push_back(a[i] + 1);
    v.push_back(a[i] - 2);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (long long i = 1; i <= n; ++i) {
    long long id = getid(a[i]), id1 = getid(a[i] + 1), id2 = getid(a[i] - 2);
    ans += a[i] * (i - 1 - (getsums(id1) - getsums(id2))) -
           (sum - (getsum(id1) - getsum(id2)));
    sum += a[i];
    add(id, a[i]);
    adds(id);
  }
  cout << setprecision(0) << fixed;
  cout << ans << endl;
  return 0;
}
