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;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
template <typename T>
static long long get(const T &freq, int idx) {
  auto pos = freq.find(idx);
  if (pos == freq.end())
    return 0;
  else
    return pos->second;
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  int N;
  cin >> N;
  vector<int> a(N);
  unordered_map<int, int> freq;
  freq.reserve(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  unsigned long long ansp = 0;
  unsigned long long ansn = 0;
  for (long long i = 0; i < N; i++) {
    long long diff = (i - (N - 1 - i)) * a[i];
    diff -= get(freq, a[i] - 1);
    diff += get(freq, a[i] + 1);
    freq[a[i]]++;
    if (diff < 0)
      ansn += -diff;
    else
      ansp += diff;
  }
  if (ansn > ansp)
    cout << "-" << ansn - ansp << '\n';
  else
    cout << ansp - ansn << '\n';
  return 0;
}
