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. 
n = int(input())
t = list(map(int, input().split()))
c = {}
u = {}
a = [0 for _ in range(n)]
for i in range(n):
    a[i] = t[i]*i - t[i]*(n-1-i);
    if t[i] not in u:
        u[t[i]] = [i]
    else:
        u[t[i]].append(i)
l = []
for x in u:
    l.append(x);
l.sort()
p = len(l)
res = 0
for i in range(p-1):
    if (l[i]+1 in u):
        l1 = len(u[l[i]])
        l2 = len(u[l[i]+1])
        j1 = 0
        j2 = 0
        t1 = 0
        t2 = 0
        for j in range(l1+l2):
            #print(j1, j2);
            #print(l1, l2)
            if (j1 == l1):
                t2 += 1
                j2 += 1
                res -= t1
            elif j2 == l2:
                t1 += 1
                j1 += 1
                res += t2
            elif (u[l[i]][j1] < u[l[i]+1][j2]):
                t1 += 1
                j1 += 1
                res += t2
            else:
                t2 += 1
                j2 += 1
                res -= t1
for i in range(n):
    res += a[i]
print(res)

