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 inf = 1e9 + 100;
const int N = 2e6 + 10;
const long long mod = 1e9 + 7;
const int base = 131;
const double pi = acos(-1);
const double eps = 1e-8;
inline long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ans;
}
unordered_map<long long, long long> mp;
long long vis[N], num[N], a[N], b[N], in[N];
long long n, m, k, x, y, z, q;
long long cx, cy, cz, cnt, sum, flag, t, ff, w;
vector<int> v[N], res;
string s;
pair<int, int> p[N];
long double ans;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    t = sum - mp[a[i] - 1] * (a[i] - 1) - mp[a[i]] * a[i] -
        mp[a[i] + 1] * (a[i] + 1);
    x = i - 1 - mp[a[i] - 1] - mp[a[i]] - mp[a[i] + 1];
    ans += (a[i] * (x)-t);
    sum += a[i];
    mp[a[i]]++;
  }
  printf("%.0LF", ans);
}
