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 + 5, mod = 1e9 + 7;
const long long base = 1000000000000000ll;
int n;
long long a[N];
map<long long, long long> mp;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  long long suf = 0, low = 0, high = 0;
  for (long long i = n; i >= 1; i--) {
    long long x = a[i];
    long long p1 = mp[x], p2 = mp[x + 1], p3 = mp[x - 1];
    long long tmp = suf - (p1 * x + p2 * (x + 1) + p3 * (x - 1));
    long long num = n - i - (p1 + p2 + p3);
    low = low + (tmp - num * x);
    while (low >= base) low -= base, high++;
    while (low < 0) low += base, high--;
    suf += x;
    mp[x]++;
  }
  if (high >= -1 && high <= 0)
    printf("%lld", high * base + low);
  else if (high > 0)
    printf("%lld%015lld\n", high, low);
  else
    printf("%lld%015lld\n", high + (low > 0), (low > 0) * base - low);
  return 0;
}
