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;
long long n, m = 0, k = 0;
bool test = false, test1 = false;
long long const N = 1e5 + 7;
long long const oo = 1000000007;
long long nChoosek(long long n, long long k) {
  if (k > n) return 0;
  if (k * 2 > n) k = n - k;
  if (k == 0) return 1;
  long long result = n;
  for (long long i = 2; i <= k; ++i) {
    result *= (n - i + 1);
    result /= i;
  }
  return result;
}
vector<long long> v, v1;
vector<pair<long long, long long> > vv;
map<long long, long long> my;
map<long long, long long> my1;
priority_queue<long long> q;
long long t[N];
long long t1[N];
long long dx[4] = {0, 1, 0, -1};
long long dy[4] = {1, 0, -1, 0};
bool in_da_prison(long long x, long long y) {
  return (x < 9 && x >= 0 && y < 9 && y >= 0);
}
long long dp(long long i) {
  if (i == 0) return 1;
  if (i < 0) return 0;
  if (t[i] != -1) return t[i];
  long long choice1 = dp(i - 1) % oo;
  long long choice2 = dp(i - k) % oo;
  return t[i] = (choice1 % oo + choice2 % oo) % oo;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i = 0, a = 0, b = 0, c = 0, d = 1e10, x = 0, y = 0, s = 0;
  string s1, s2, s3, s4, s5;
  long double j = 0.0;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a;
    my[a]++;
    j -= (my[a - 1] - my[a + 1]);
    j += a * i - a * (n - i - 1);
  }
  cout << fixed << setprecision(0) << j;
  return 0;
}
