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. 
if __name__ == '__main__':
	n=int(input())

	soujanya={}

	sum=0
	k=input().split()
	a={}
	for j in range(n):
		i=j+1
		a[i]=int(k[j])
		sum+=a[i]
		if a[i] in soujanya.keys():
			soujanya[a[i]]+=1
		else:
			soujanya[a[i]]=1
		

	counter=n
	ans=0		
	for j in range(n):
		i=j+1
		temp=counter
		if (a[i]-1) in soujanya.keys():
			k1=soujanya[a[i]-1]
		else:
			k1=0
		
		if (a[i]) in soujanya.keys():
			k2=soujanya[a[i]]
		else:
			k2=0
			
		if (a[i]+1) in soujanya.keys():
			k3=soujanya[a[i]+1]
		else:
			k3=0
		
		temp-=k1+k2+k3
		k=sum-(k1*(a[i]-1)+k2*a[i]+k3*(a[i]+1))
		ans+=k-temp*a[i]
		counter=counter-1
		sum-=a[i]
		soujanya[a[i]]-=1
		
		

	print(ans)	