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 bit[600010], bit1[600010];
int sz;
long long a[200010], b[600010];
void update(int x, int val) {
  while (x <= sz) {
    bit[x] += 1ll * val;
    x += (x) & (-x);
  }
}
long long sum(int x) {
  long long res = 0;
  while (x > 0) {
    res += bit[x];
    x -= (x) & (-x);
  }
  return res;
}
void update1(int x) {
  while (x <= sz) {
    bit1[x]++;
    x += (x) & (-x);
  }
}
long long sum1(int x) {
  long long res = 0;
  while (x > 0) {
    res += bit1[x];
    x -= (x) & (-x);
  }
  return res;
}
int id(int x) { return lower_bound(b + 1, b + sz + 1, x) - b; }
int main() {
  int n;
  scanf("%d", &n);
  sz = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[++sz] = a[i];
    b[++sz] = a[i] - 2;
    b[++sz] = a[i] + 1;
  }
  sort(b + 1, b + sz + 1);
  int nn = unique(b + 1, b + sz + 1) - b - 1;
  sz = nn;
  long double ans = 0;
  long long tol = 0;
  for (int i = 1; i <= n; i++) {
    tol += a[i];
    int iid = id(a[i]);
    update(iid, a[i]);
    update1(iid);
    long long res = tol - (sum(id(a[i] + 1)) - sum(id(a[i] - 2)));
    long long res1 = i - (sum1(id(a[i] + 1)) - sum1(id(a[i] - 2)));
    ans += res1 * a[i] - res;
  }
  cout << fixed << setprecision(0) << ans;
}
