#include <bits/stdc++.h>
using namespace std;
long long n;
long long st[5 * 1000005], pre[1000005];
map<long long, long long> mp;
long long query(long long p, long long l, long long r, long long i,
                long long j) {
  if (l > j || r < i) return 0;
  if (i <= l and r <= j) return st[p];
  long long p1 = query(2 * p, l, (l + r) / 2, i, j);
  long long p2 = query(2 * p + 1, (l + r) / 2 + 1, r, i, j);
  return p1 + p2;
}
void update(long long p, long long l, long long r, long long i) {
  if (i < l || i > r) return;
  if (l == r) {
    st[p] = 1;
    return;
  }
  update(2 * p, l, (l + r) / 2, i);
  update(2 * p + 1, (l + r) / 2 + 1, r, i);
  st[p] = st[2 * p] + st[2 * p + 1];
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    mp[x] = i;
  }
  long long i = 1;
  auto it = mp.end();
  it--;
  for (long long i = 0; i < mp.size(); i++) {
    pre[it->second] = query(1, 0, n - 1, 0, it->second);
    update(1, 0, n - 1, it->second);
    it--;
  }
  for (long long i = 0; i < 5 * 1000005; i++) st[i] = 0;
  long long ans = 0;
  for (auto it = mp.begin(); it != mp.end(); it++) {
    long long x = query(1, 0, n - 1, it->second, n - 1);
    update(1, 0, n - 1, it->second);
    ans += (pre[it->second] * x);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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;
long long n;
long long st[5 * 1000005], pre[1000005];
map<long long, long long> mp;
long long query(long long p, long long l, long long r, long long i,
                long long j) {
  if (l > j || r < i) return 0;
  if (i <= l and r <= j) return st[p];
  long long p1 = query(2 * p, l, (l + r) / 2, i, j);
  long long p2 = query(2 * p + 1, (l + r) / 2 + 1, r, i, j);
  return p1 + p2;
}
void update(long long p, long long l, long long r, long long i) {
  if (i < l || i > r) return;
  if (l == r) {
    st[p] = 1;
    return;
  }
  update(2 * p, l, (l + r) / 2, i);
  update(2 * p + 1, (l + r) / 2 + 1, r, i);
  st[p] = st[2 * p] + st[2 * p + 1];
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    mp[x] = i;
  }
  long long i = 1;
  auto it = mp.end();
  it--;
  for (long long i = 0; i < mp.size(); i++) {
    pre[it->second] = query(1, 0, n - 1, 0, it->second);
    update(1, 0, n - 1, it->second);
    it--;
  }
  for (long long i = 0; i < 5 * 1000005; i++) st[i] = 0;
  long long ans = 0;
  for (auto it = mp.begin(); it != mp.end(); it++) {
    long long x = query(1, 0, n - 1, it->second, n - 1);
    update(1, 0, n - 1, it->second);
    ans += (pre[it->second] * x);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```