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>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int N = 2e5 + 7;
const int M = 1e6 + 7;
vector<long long> vec, fb, fc;
long long a[N];
long long vis[M];
long long dp[N];
int getVal(long long first) {
  return lower_bound(vec.begin(), vec.end(), first) - vec.begin();
}
signed main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    vec.push_back(a[i] - 1);
    vec.push_back(a[i]);
    vec.push_back(a[i] + 1);
  }
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  long long sum = 0;
  unsigned long long ans_1 = 0, ans_2 = 0;
  for (int i = 1; i <= n; i++) {
    sum += a[i];
    int x1 = getVal(a[i] - 1), x2 = getVal(a[i]), x3 = getVal(a[i] + 1);
    vis[x2]++;
    dp[i] = a[i] * (long long)i + vis[x3] - vis[x1] - sum;
    if (dp[i] < 0)
      ans_1 += (unsigned long long)(-dp[i]);
    else
      ans_2 += (unsigned long long)dp[i];
  }
  if (ans_1 > ans_2) {
    cout << "-" << ans_1 - ans_2 << "\n";
  } else
    cout << ans_2 - ans_1 << "\n";
  return 0;
}
