#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
using ii = pair<long long int, long long int>;
class prioritize {
 public:
  bool operator()(ii &p1, ii &p2) { return p1.first < p2.first; }
};
auto start_time = high_resolution_clock::now();
void time() { return; }
void ojudge() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  return;
}
long long int n;
vector<long long int> a;
using vi = vector<long long int>;
vi tree[4000004];
void build(long long int curr, long long int tl, long long int tr) {
  if (tl == tr) {
    tree[curr].push_back(a[tl]);
    return;
  }
  long long int mid = (tl + tr) / 2;
  build(2 * curr, tl, mid);
  build(2 * curr + 1, mid + 1, tr);
  long long int sz1 = mid - tl + 1, sz2 = tr - (mid + 1) + 1;
  long long int i = 0, j = 0;
  while (i < sz1 && j < sz2) {
    if (tree[2 * curr][i] <= tree[2 * curr + 1][j]) {
      tree[curr].push_back(tree[2 * curr][i]);
      i++;
    } else {
      tree[curr].push_back(tree[2 * curr + 1][j]);
      j++;
    }
  }
  while (i < sz1) {
    tree[curr].push_back(tree[2 * curr][i]);
    i++;
  }
  while (j < sz2) {
    tree[curr].push_back(tree[2 * curr + 1][j]);
    j++;
  }
  return;
}
long long int Query_greater(long long int curr, long long int tl,
                            long long int tr, long long int ql,
                            long long int qr, long long int val) {
  if (tl > qr || tr < ql) return 0;
  if (tl >= ql && tr <= qr) {
    auto it = upper_bound((tree[curr]).begin(), (tree[curr]).end(), val);
    if (it == tree[curr].end()) return 0;
    long long int id = it - tree[curr].begin();
    return (tr - tl + 1 - id);
  }
  long long int mid = (tl + tr) / 2;
  return Query_greater(2 * curr, tl, mid, ql, qr, val) +
         Query_greater(2 * curr + 1, mid + 1, tr, ql, qr, val);
}
long long int Query_smaller(long long int curr, long long int tl,
                            long long int tr, long long int ql,
                            long long int qr, long long int val) {
  if (tl > qr || tr < ql) return 0;
  if (tl >= ql && tr <= qr) {
    auto it = upper_bound((tree[curr]).begin(), (tree[curr]).end(), val - 1);
    long long int id = it - tree[curr].begin();
    return id;
  }
  long long int mid = (tl + tr) / 2;
  return Query_smaller(2 * curr, tl, mid, ql, qr, val) +
         Query_smaller(2 * curr + 1, mid + 1, tr, ql, qr, val);
}
int main() {
  ojudge();
  cin >> n;
  a.resize(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  build(1, 0, n - 1);
  long long int ans = 0;
  for (long long int i = 1; i < n - 1; i++) {
    long long int left = Query_greater(1, 0, n - 1, 0, i - 1, a[i]);
    long long int right = Query_smaller(1, 0, n - 1, i + 1, n - 1, a[i]);
    ans += left * right;
  }
  cout << ans << "\n";
  time();
  return 0;
}

### Prompt

Your challenge is to write 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;
using namespace std::chrono;
using ii = pair<long long int, long long int>;
class prioritize {
 public:
  bool operator()(ii &p1, ii &p2) { return p1.first < p2.first; }
};
auto start_time = high_resolution_clock::now();
void time() { return; }
void ojudge() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  return;
}
long long int n;
vector<long long int> a;
using vi = vector<long long int>;
vi tree[4000004];
void build(long long int curr, long long int tl, long long int tr) {
  if (tl == tr) {
    tree[curr].push_back(a[tl]);
    return;
  }
  long long int mid = (tl + tr) / 2;
  build(2 * curr, tl, mid);
  build(2 * curr + 1, mid + 1, tr);
  long long int sz1 = mid - tl + 1, sz2 = tr - (mid + 1) + 1;
  long long int i = 0, j = 0;
  while (i < sz1 && j < sz2) {
    if (tree[2 * curr][i] <= tree[2 * curr + 1][j]) {
      tree[curr].push_back(tree[2 * curr][i]);
      i++;
    } else {
      tree[curr].push_back(tree[2 * curr + 1][j]);
      j++;
    }
  }
  while (i < sz1) {
    tree[curr].push_back(tree[2 * curr][i]);
    i++;
  }
  while (j < sz2) {
    tree[curr].push_back(tree[2 * curr + 1][j]);
    j++;
  }
  return;
}
long long int Query_greater(long long int curr, long long int tl,
                            long long int tr, long long int ql,
                            long long int qr, long long int val) {
  if (tl > qr || tr < ql) return 0;
  if (tl >= ql && tr <= qr) {
    auto it = upper_bound((tree[curr]).begin(), (tree[curr]).end(), val);
    if (it == tree[curr].end()) return 0;
    long long int id = it - tree[curr].begin();
    return (tr - tl + 1 - id);
  }
  long long int mid = (tl + tr) / 2;
  return Query_greater(2 * curr, tl, mid, ql, qr, val) +
         Query_greater(2 * curr + 1, mid + 1, tr, ql, qr, val);
}
long long int Query_smaller(long long int curr, long long int tl,
                            long long int tr, long long int ql,
                            long long int qr, long long int val) {
  if (tl > qr || tr < ql) return 0;
  if (tl >= ql && tr <= qr) {
    auto it = upper_bound((tree[curr]).begin(), (tree[curr]).end(), val - 1);
    long long int id = it - tree[curr].begin();
    return id;
  }
  long long int mid = (tl + tr) / 2;
  return Query_smaller(2 * curr, tl, mid, ql, qr, val) +
         Query_smaller(2 * curr + 1, mid + 1, tr, ql, qr, val);
}
int main() {
  ojudge();
  cin >> n;
  a.resize(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  build(1, 0, n - 1);
  long long int ans = 0;
  for (long long int i = 1; i < n - 1; i++) {
    long long int left = Query_greater(1, 0, n - 1, 0, i - 1, a[i]);
    long long int right = Query_smaller(1, 0, n - 1, i + 1, n - 1, a[i]);
    ans += left * right;
  }
  cout << ans << "\n";
  time();
  return 0;
}

```