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 = 400000 + 5;
int n, a[N], cnt[N];
long long sum[N];
map<int, int> id;
struct Num {
  int len;
  int d[100];
  Num() {
    len = 0;
    memset(d, 0, sizeof d);
  }
  Num(long long a) {
    len = 0;
    memset(d, 0, sizeof d);
    while (a) {
      d[++len] = a % 10;
      a /= 10;
    }
  }
  void Print() {
    for (int i = len; i >= 1; i--) printf("%d", d[i]);
  }
};
Num operator+(const Num &A, const Num &B) {
  Num ret;
  ret.len = max(A.len, B.len);
  for (int i = 1; i <= ret.len; i++) {
    ret.d[i] += A.d[i] + B.d[i];
    ret.d[i + 1] += ret.d[i] / 10;
    ret.d[i] %= 10;
  }
  if (ret.d[ret.len + 1]) ++ret.len;
  return ret;
}
Num operator-(const Num &A, const Num &B) {
  Num ret;
  ret.len = max(A.len, B.len) + 1;
  for (int i = 1; i <= ret.len; i++) {
    ret.d[i] += A.d[i] - B.d[i];
    if (ret.d[i] < 0) {
      ret.d[i] += 10;
      ret.d[i + 1]--;
    }
  }
  while (!ret.d[ret.len] && ret.len > 1) --ret.len;
  return ret;
}
bool operator>(const Num &A, const Num &B) {
  if (A.len > B.len) return 1;
  if (A.len < B.len) return 0;
  for (int i = A.len; i >= 1; i--) {
    if (A.d[i] < B.d[i]) return 0;
    if (A.d[i] > B.d[i]) return 1;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int l = 0;
  for (int i = 1; i <= n; i++) {
    if (!id[a[i]]) id[a[i]] = ++l;
  }
  Num ans1 = 0;
  Num ans2 = 0;
  for (int i = n; i >= 1; i--) {
    ans1 = ans1 + sum[i + 1];
    ans1 = ans1 + cnt[id[a[i] - 1]];
    ans2 = ans2 + 1LL * a[i] * (n - i);
    ans2 = ans2 + cnt[id[a[i] + 1]];
    cnt[id[a[i]]]++;
    sum[i] = sum[i + 1] + a[i];
  }
  Num ans = 0;
  if (ans1 > ans2) {
    ans = ans1 - ans2;
    ans.Print();
  } else {
    printf("-");
    ans = ans2 - ans1;
    ans.Print();
  }
  return 0;
}
