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;
int a[200002], p[200002];
unsigned long long s[200002];
bool cmp(int x, int y) {
  if (a[x] == a[y]) return x < y;
  return a[x] < a[y];
}
int main() {
  int n, i, j, t1, q, t2, t, f, ti, tti, tj, ii;
  unsigned long long ans1 = 0, ans2 = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", a + i);
    p[i] = i;
  }
  sort(p + 1, p + n + 1, cmp);
  for (i = 2, s[j = 1] = (unsigned long long)a[1]; i <= n; j = i++)
    s[i] = s[j] + (unsigned long long)a[i];
  t = p[1];
  ans1 += (unsigned long long)t * (unsigned long long)a[t], ans2 += s[t];
  for (i = 2, t1 = ii = 1; i <= n && a[p[i]] == a[p[ii]]; ii = i++) {
    t = p[i];
    ans1 += (unsigned long long)t * (unsigned long long)a[t], ans2 += s[t];
  }
  for (t2 = i; i <= n; i++) {
    if (a[p[t1]] + 1 != a[p[t2]]) {
      ti = t2;
      do {
        t = p[ti];
        ans1 += (unsigned long long)t * (unsigned long long)a[t], ans2 += s[t];
        if (ans1 > ans2)
          ans1 -= ans2, ans2 = 0;
        else
          ans2 -= ans1, ans1 = 0;
        tti = ti++;
      } while (a[p[ti]] == a[p[tti]]);
      t1 = t2, t2 = ti, i = ti - 1;
    } else {
      ti = t2, tj = t1;
      do {
        t = p[ti];
        ans1 += (unsigned long long)t * (unsigned long long)a[t], ans2 += s[t];
        if (ans1 > ans2)
          ans1 -= ans2, ans2 = 0;
        else
          ans2 -= ans1, ans1 = 0;
        for (; tj < t2 && p[ti] > p[tj]; tj++)
          ;
        ans2 += (unsigned long long)tj - (unsigned long long)t1;
        ans1 += (unsigned long long)t2 - (unsigned long long)tj;
        tti = ti++;
      } while (a[p[ti]] == a[p[tti]]);
      t1 = t2, t2 = ti, i = ti - 1;
    }
  }
  if (ans1 >= ans2)
    printf("%llu\n", ans1 - ans2);
  else
    printf("-%llu\n", ans2 - ans1);
  return 0;
}
