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>
template <typename T>
std::string to_string(T value) {
  std::ostringstream os;
  os << value;
  return os.str();
}
using namespace std;
long double cnt[800005];
long double sm[800005];
void add(int node, int b, int e, int val, long long r_val) {
  if (b > val || e < val) {
    return;
  }
  if (b <= val && val <= e) {
    cnt[node]++;
    sm[node] += r_val;
    if (b == e) {
      return;
    }
  }
  int md = (b + e) / 2;
  int left = 2 * node;
  int right = left + 1;
  add(left, b, md, val, r_val);
  add(right, md + 1, e, val, r_val);
  return;
}
long double ans;
void query(int node, int b, int e, int x, int y, int r_val) {
  if (b > y || e < x) {
    return;
  }
  if (x <= b && e <= y) {
    long long t = cnt[node] * r_val;
    t -= sm[node];
    ans += t;
    return;
  }
  int md = (b + e) / 2;
  int left = 2 * node;
  int right = left + 1;
  query(left, b, md, x, y, r_val);
  query(right, md + 1, e, x, y, r_val);
  return;
}
int main() {
  int n;
  scanf("%d", &n);
  int ns[n];
  map<int, int> mp1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &ns[i]);
    mp1.insert({ns[i], 0});
  }
  int val = 0;
  for (auto it = mp1.begin(); it != mp1.end(); it++) {
    val++;
    it->second = val;
  }
  int mx = val;
  map<int, int> mp2;
  for (auto it = mp1.begin(); it != mp1.end(); it++) {
    mp2.insert({it->second, it->first});
  }
  add(1, 1, mx, mp1[ns[0]], ns[0]);
  ans = 0;
  map<int, int> mp;
  mp.insert({ns[0], mp1[ns[0]]});
  for (int i = 1; i < n; i++) {
    int val = mp1[ns[i]];
    long long r_val = ns[i];
    auto it = mp.lower_bound(r_val + 2);
    if (it != mp.end()) {
      query(1, 1, mx, it->second, mx, r_val);
    }
    it = mp.lower_bound(r_val - 1);
    if (it != mp.begin()) {
      it--;
      query(1, 1, mx, 1, it->second, r_val);
    }
    add(1, 1, mx, val, r_val);
    mp.insert({r_val, val});
  }
  string s = to_string(ans);
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '.') {
      break;
    }
    cout << s[i];
  }
  cout << endl;
  return 0;
}
