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 int maxn = 1e6 + 5;
long double arr[maxn];
vector<long double> v;
map<long double, long double> mp;
long double tree[maxn], tree1[maxn];
void update(int idx, long double val) {
  while (idx < maxn) {
    tree[idx] += val;
    idx += (idx & (-idx));
  }
}
void update1(int idx, long double val) {
  while (idx < maxn) {
    tree1[idx] += val;
    idx += (idx & (-idx));
  }
}
long double query(int idx) {
  long double ret = 0;
  while (idx) {
    ret += tree[idx];
    idx -= (idx & (-idx));
  }
  return ret;
}
long double query1(int idx) {
  long double ret = 0;
  while (idx) {
    ret += tree1[idx];
    idx -= (idx & (-idx));
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i;
  long double n;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
    v.push_back(arr[i]);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  int cnt = 2;
  mp[v[0]] = 2;
  for (i = 1; i < int(v.size()); i++) {
    if (v[i] - v[i - 1] > 1) {
      ++cnt;
      mp[v[i]] = ++cnt;
    } else {
      mp[v[i]] = ++cnt;
    }
  }
  long double ans = 0;
  long double tot = 0;
  for (i = n; i > 0; i--) {
    long double sumlow = query(mp[arr[i]] - 2);
    long double cntlow = query1(mp[arr[i]] - 2);
    long double sumhigh = tot - query(mp[arr[i]] + 1);
    long double cnthigh = n - i - query1(mp[arr[i]] + 1);
    ans += sumlow - cntlow * arr[i];
    ans += sumhigh - cnthigh * arr[i];
    update(mp[arr[i]], arr[i]);
    update1(mp[arr[i]], 1);
    tot += arr[i];
  }
  cout << setprecision(30) << ans;
  return 0;
}
