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 N = 2e5 + 10;
const int mod = 1e9 + 7;
int n;
int a[N];
map<int, int> m;
long long sum[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    m[a[i]]++;
  }
  sum[n] = a[n];
  for (int i = n - 1; i >= 1; i--) sum[i] = sum[i + 1] + a[i];
  long long a1 = 0;
  long long a2 = 0;
  for (int i = 1; i <= n; i++) {
    m[a[i]]--;
    a1 += sum[i + 1] - 1ll * a[i] * (n - i);
    a1 -= m[a[i] + 1];
    a1 += m[a[i] - 1];
    a2 += a1 / mod;
    a1 = a1 % mod;
  }
  if (a2 < 0) {
    printf("-");
    a2 = -1 * a2;
    unsigned long long ans = a2;
    ans *= mod;
    a1 = -1 * a1;
    ans += a1;
    printf("%llu\n", ans);
  } else if (a2 == 0)
    printf("%lld\n", a1);
  else {
    unsigned long long ans = a2;
    ans *= mod;
    ans += a1;
    printf("%llu\n", ans);
  }
}
