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 int mod = 1e9 + 7;
const long long inf = 3e18 + 5;
int add(int a, int b) { return (a += b) < mod ? a : a - mod; }
int mul(int a, int b) { return 1LL * a * b % mod; }
int sub(int a, int b) { return (a -= b) < 0 ? a + mod : a; }
int ctz(int x) { return __builtin_ctz(x); }
int clz(int x) { return __builtin_clz(x); }
const int maxn = 2e5 + 5;
long long ft[maxn][2];
void up(int i, int val, int t) {
  while (i < maxn) {
    ft[i][t] += val;
    i += i & (-i);
  }
}
long double qry(int i, int t) {
  long long ret = 0;
  while (i > 0) {
    ret += ft[i][t];
    i -= i & (-i);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  map<int, int> has;
  vector<int> v, a(n);
  for (int& x : a) {
    cin >> x;
    has[x]++;
    v.push_back(x);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  vector<int> mapped(n);
  for (int i = 0; i < n; i++) {
    mapped[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  }
  long double res = 0;
  for (int i = 0; i < n; i++) {
    long double freqleft =
        qry((has[a[i] - 1] ? mapped[i] - 2 : mapped[i] - 1), 0);
    long double sumleft =
        qry((has[a[i] - 1] ? mapped[i] - 2 : mapped[i] - 1), 1);
    long double freqright =
        qry(n, 0) - qry((has[a[i] + 1] ? mapped[i] + 2 : mapped[i] + 1) - 1, 0);
    long double sumright =
        qry(n, 1) - qry((has[a[i] + 1] ? mapped[i] + 2 : mapped[i] + 1) - 1, 1);
    res += freqleft * a[i] - sumleft;
    res -= sumright - freqright * a[i];
    up(mapped[i], +1, 0);
    up(mapped[i], a[i], 1);
  }
  cout << fixed << setprecision(0) << res;
}
