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 MD = 1000000000000000000;
int main() {
  int n;
  cin >> n;
  long long ans = 0, sum = 0;
  long long carry = 0;
  unordered_map<int, int> mp;
  for (int j = 0; j < n; j++) {
    int a;
    cin >> a;
    ans += (long long)a * j - sum;
    ans -= mp[a - 1];
    ans += mp[a + 1];
    if (abs(ans) >= MD) {
      carry += ans / MD;
      ans %= MD;
    }
    mp[a]++;
    sum += a;
  }
  if (carry > 0 && ans < 0) {
    carry--;
    ans += MD;
  } else if (carry < 0 && ans > 0) {
    carry++;
    ans -= MD;
  }
  if (carry == 0)
    printf("%lld\n", ans);
  else {
    if (carry < 0) {
      carry = -carry;
      ans = -ans;
      printf("-");
    }
    printf("%lld%.18lld\n", carry, ans);
  }
  return 0;
}
