Let's denote a function 

<image>

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 = 2e5 + 10;
const int TN = 4e5 + 10;
int a[N];
long long bits[TN];
long long cnt[TN];
void insert(int x, long long v) {
  while (x < TN) {
    bits[x] += v;
    cnt[x]++;
    x += x & (-x);
  }
}
pair<long long, long long> query(int x) {
  pair<long long, long long> res = make_pair(0, 0);
  while (x) {
    res.first += bits[x];
    res.second += cnt[x];
    x -= x & (-x);
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  long long sum = 0;
  vector<int> fs;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    fs.push_back(a[i]);
  }
  sort(fs.begin(), fs.end());
  int idx = 1;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    if (i == 0 || fs[i] != fs[i - 1]) {
      mp[fs[i]] = idx;
      idx++;
    }
  }
  long long ans = 0;
  long long MOD = 10000000000LL;
  long long top = 0;
  for (int i = 0; i < n; i++) {
    sum += a[i];
    insert(mp[a[i]], a[i]);
    int pos = mp[a[i]];
    pair<long long, long long> res;
    if (pos > 0) {
      if (mp.count(a[i] - 1) == 0) {
        res = query(mp[a[i]] - 1);
        ans += a[i] * res.second - res.first;
        top += ans / MOD;
        ans %= MOD;
      } else if (pos > 1) {
        res = query(mp[a[i]] - 2);
        ans += a[i] * res.second - res.first;
        top += ans / MOD;
        ans %= MOD;
      }
    }
    if (mp.count(a[i] + 1) == 0) {
      res = query(mp[a[i]]);
      ans += a[i] * (i + 1 - res.second) - (sum - res.first);
      top += ans / MOD;
      ans %= MOD;
    } else {
      res = query(mp[a[i]] + 1);
      ans += a[i] * (i + 1 - res.second) - (sum - res.first);
      top += ans / MOD;
      ans %= MOD;
    }
  }
  if (top == 0)
    cout << ans << endl;
  else {
    if (top < 0) {
      top = -top;
      ans = -ans;
      printf("-");
    }
    if (ans >= 0)
      printf("%I64d%010I64d\n", top, ans);
    else if (top == 1) {
      cout << MOD + ans << endl;
    } else {
      printf("%I64d%010I64d\n", top - 1, MOD + ans);
    }
  }
}
