#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using ll = long long;
const int maxn = 1e6;
const int inf = 1e9;
const int mod = 1e9 + 7;
void open_io() {
  freopen("in", "r", stdin);
  freopen("out", "w", stdout);
}
void fast_io() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
}
long long t[4 * maxn + 15], a[maxn + 5], d[maxn + 5], n, ans;
void update(long long vi, long long vl, long long vr, long long pos) {
  if (vl == vr and vl == pos) {
    t[vi] = 1;
    return;
  }
  long long mid = (vl + vr) / 2;
  if (pos <= mid)
    update(2 * vi, vl, mid, pos);
  else
    update(2 * vi + 1, mid + 1, vr, pos);
  t[vi] = t[2 * vi] + t[2 * vi + 1];
}
long long query(long long vi, long long vl, long long vr, long long l,
                long long r) {
  if (l > r) return 0;
  if (vl == l and vr == r) return t[vi];
  long long mid = (vl + vr) / 2;
  return query(2 * vi, vl, mid, l, min(mid, r)) +
         query(2 * vi + 1, mid + 1, vr, max(l, mid + 1), r);
}
void compress() {
  for (long long i = 1; i <= n; i++) d[i] = a[i];
  sort(d + 1, d + n + 1);
  for (long long i = 1; i <= n; i++)
    a[i] = lower_bound(d + 1, d + n + 1, a[i]) - d;
}
int32_t main() {
  fast_io();
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  compress();
  for (long long i = 1; i <= n; i++) {
    long long lft = query(1, 1, n, a[i] + 1, n);
    long long rgt = (a[i] - 1) - (i - 1 - lft);
    ans += lft * rgt;
    update(1, 1, n, a[i]);
  }
  cout << ans << '\n';
}

### Prompt

Develop a solution in Cpp to the problem described below:

The Romans have attacked again. This time they are much more than the Persians but Shapur is ready to defeat them. He says: "A lion is never afraid of a hundred sheep". 

Nevertheless Shapur has to find weaknesses in the Roman army to defeat them. So he gives the army a weakness number.

In Shapur's opinion the weakness of an army is equal to the number of triplets i, j, k such that i < j < k and ai > aj > ak where ax is the power of man standing at position x. The Roman army has one special trait — powers of all the people in it are distinct.

Help Shapur find out how weak the Romans are.

Input

The first line of input contains a single number n (3 ≤ n ≤ 106) — the number of men in Roman army. Next line contains n different positive integers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 109) — powers of men in the Roman army. 

Output

A single integer number, the weakness of the Roman army. 

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

3
3 2 1


Output

1


Input

3
2 3 1


Output

0


Input

4
10 8 3 1


Output

4


Input

4
1 5 4 3


Output

1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using ll = long long;
const int maxn = 1e6;
const int inf = 1e9;
const int mod = 1e9 + 7;
void open_io() {
  freopen("in", "r", stdin);
  freopen("out", "w", stdout);
}
void fast_io() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
}
long long t[4 * maxn + 15], a[maxn + 5], d[maxn + 5], n, ans;
void update(long long vi, long long vl, long long vr, long long pos) {
  if (vl == vr and vl == pos) {
    t[vi] = 1;
    return;
  }
  long long mid = (vl + vr) / 2;
  if (pos <= mid)
    update(2 * vi, vl, mid, pos);
  else
    update(2 * vi + 1, mid + 1, vr, pos);
  t[vi] = t[2 * vi] + t[2 * vi + 1];
}
long long query(long long vi, long long vl, long long vr, long long l,
                long long r) {
  if (l > r) return 0;
  if (vl == l and vr == r) return t[vi];
  long long mid = (vl + vr) / 2;
  return query(2 * vi, vl, mid, l, min(mid, r)) +
         query(2 * vi + 1, mid + 1, vr, max(l, mid + 1), r);
}
void compress() {
  for (long long i = 1; i <= n; i++) d[i] = a[i];
  sort(d + 1, d + n + 1);
  for (long long i = 1; i <= n; i++)
    a[i] = lower_bound(d + 1, d + n + 1, a[i]) - d;
}
int32_t main() {
  fast_io();
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  compress();
  for (long long i = 1; i <= n; i++) {
    long long lft = query(1, 1, n, a[i] + 1, n);
    long long rgt = (a[i] - 1) - (i - 1 - lft);
    ans += lft * rgt;
    update(1, 1, n, a[i]);
  }
  cout << ans << '\n';
}

```