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;
struct i128 {
  static const long long INF = 1e18;
  long long h, l;
  i128() {}
  i128(long long _h, long long _l) : h(_h), l(_l) {
    if (l >= INF) {
      l -= INF;
      h++;
    } else if (l <= -INF) {
      l += INF;
      h--;
    } else if (h > 0 && l < 0) {
      l += INF;
      h--;
    } else if (h < 0 && l > 0) {
      l -= INF;
      h++;
    }
  }
  i128 operator+(const i128 b) { return i128(h + b.h, l + b.l); }
  i128 operator-(const i128 b) { return i128(h - b.h, l - b.l); }
  void out() {
    if (h == 0)
      printf("%lld", l);
    else
      printf("%lld%018lld", h, abs(l));
  }
};
const int N = 2e5 + 7;
int n;
void solve() {
  map<int, int> M;
  i128 ans(0, 0);
  for (int i = (1); i < (n + 1); ++i) {
    int a;
    scanf("%d", &a);
    ans = ans + i128(0, (2ll * i - n - 1) * a);
    if (M.count(a - 1)) ans = ans - i128(0, M[a - 1]);
    if (M.count(a + 1)) ans = ans + i128(0, M[a + 1]);
    M[a]++;
  }
  ans.out();
  puts("");
}
int main() {
  while (~scanf("%d", &n)) solve();
  return 0;
}
