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 inf = 0x3f3f3f3f;
const int maxn = 200000 + 100;
int n;
int a[maxn];
vector<int> disc;
long long bit1[maxn];
long long bitsum[maxn];
int inline lsb(int x) { return x & (-x); }
long long query(long long cc[], int x) {
  long long ret = 0;
  while (x) {
    ret += cc[x];
    x -= lsb(x);
  }
  return ret;
}
void add(long long cc[], int x, int b) {
  while (x < disc.size()) {
    cc[x] += b;
    x += lsb(x);
  }
}
int main() {
  while (~scanf("%d", &n)) {
    disc = {-inf, inf};
    for (int i = 0; i < n; i++) {
      scanf("%d", a + i);
      disc.push_back(a[i]);
    }
    sort(disc.begin(), disc.end());
    disc.erase(unique(disc.begin(), disc.end()), disc.end());
    long double ans = 0;
    for (int i = 0; i < n; i++) {
      int id = lower_bound(disc.begin(), disc.end(), a[i]) - disc.begin();
      int L = id - 1, R = id;
      if (disc[id - 1] == a[i] - 1) L--;
      if (disc[id + 1] == a[i] + 1) R++;
      ans += a[i] * (query(bit1, disc.size() - 1) - query(bit1, R) +
                     query(bit1, L)) -
             (query(bitsum, disc.size() - 1) - query(bitsum, R) +
              query(bitsum, L));
      add(bit1, id, 1);
      add(bitsum, id, a[i]);
    }
    cout << fixed << setprecision(0) << ans << '\n';
  }
}
