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;
int Test = 1;
const int N = 2e5 + 10;
const int MOD = 1e9 + 7;
struct FenwickTree {
  vector<pair<long long, int> > tree;
  int n;
  void init(int _n) {
    n = _n;
    tree.assign(n + 10, {0, 0});
  }
  void update(int x, int val) {
    for (; x <= n; x += x & -x) {
      tree[x].first += val;
      tree[x].second++;
    }
  }
  pair<long long, int> get(int x) {
    pair<long long, int> ret = {0, 0};
    for (; x; x -= x & -x) {
      ret.first += tree[x].first;
      ret.second += tree[x].second;
    }
    return ret;
  }
} BIT;
long long prefix[N];
pair<int, int> a[N];
int n;
int main() {
  if (fopen("903D"
            ".inp",
            "r")) {
    freopen(
        "903D"
        ".inp",
        "r", stdin);
    freopen(
        "903D"
        ".out",
        "w", stdout);
  }
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a + 1, a + 1 + n);
  BIT.init(n);
  int j = 1;
  long double ans = 0;
  for (int i = 1; i <= n; i++) {
    prefix[i] = prefix[i - 1] + a[i].first;
    while (a[i].first - a[j].first > 1) {
      BIT.update(a[j].second, a[j].first);
      j++;
    }
    pair<long long, int> p = BIT.get(a[i].second);
    ans += 1ll * a[i].first * p.second - p.first;
    ans += prefix[j - 1] - p.first - 1ll * a[i].first * (j - 1 - p.second);
  }
  cout << fixed << setprecision(0) << ans;
}
