#include <bits/stdc++.h>
using namespace std;
long long Seg[3][1 << 21 + 2], X, Cnt = 1;
pair<long long, int> P[1000001];
int l_q, r_q, Len;
long long RSQ(int l, int r, int root) {
  if (l > r_q || l_q > r || l > r) return 0;
  if (l >= l_q && r <= r_q) return Seg[Len][root];
  int mid = (r + l) / 2;
  long long C1 = RSQ(l, mid, root * 2);
  long long C2 = RSQ(mid + 1, r, root * 2 + 1);
  return C1 + C2;
}
void Update(int l, int r, int root) {
  if (l > r_q || l_q > r || l > r) return;
  if (l >= l_q && r <= r_q) {
    Seg[Len][root] += X;
    return;
  }
  int mid = (r + l) / 2;
  Update(l, mid, root * 2);
  Update(mid + 1, r, root * 2 + 1);
  Seg[Len][root] = Seg[Len][root * 2] + Seg[Len][root * 2 + 1];
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> P[i].first, P[i].second = i;
  sort(P, P + n);
  for (int i = 0; i < n; i++) {
    P[P[i].second].first = Cnt++;
  }
  reverse(P, P + n);
  for (int i = 0; i < n; i++) {
    for (int L = 0; L < 3; L++) {
      l_q = 1;
      r_q = P[i].first - 1;
      Len = L - 1;
      long long val;
      if (L == 0)
        val = 1;
      else
        val = RSQ(1, n, 1);
      l_q = r_q = P[i].first;
      Len = L;
      X = val;
      Update(1, n, 1);
    }
  }
  cout << Seg[2][1] << endl;
  return 0;
}

### 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;
long long Seg[3][1 << 21 + 2], X, Cnt = 1;
pair<long long, int> P[1000001];
int l_q, r_q, Len;
long long RSQ(int l, int r, int root) {
  if (l > r_q || l_q > r || l > r) return 0;
  if (l >= l_q && r <= r_q) return Seg[Len][root];
  int mid = (r + l) / 2;
  long long C1 = RSQ(l, mid, root * 2);
  long long C2 = RSQ(mid + 1, r, root * 2 + 1);
  return C1 + C2;
}
void Update(int l, int r, int root) {
  if (l > r_q || l_q > r || l > r) return;
  if (l >= l_q && r <= r_q) {
    Seg[Len][root] += X;
    return;
  }
  int mid = (r + l) / 2;
  Update(l, mid, root * 2);
  Update(mid + 1, r, root * 2 + 1);
  Seg[Len][root] = Seg[Len][root * 2] + Seg[Len][root * 2 + 1];
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> P[i].first, P[i].second = i;
  sort(P, P + n);
  for (int i = 0; i < n; i++) {
    P[P[i].second].first = Cnt++;
  }
  reverse(P, P + n);
  for (int i = 0; i < n; i++) {
    for (int L = 0; L < 3; L++) {
      l_q = 1;
      r_q = P[i].first - 1;
      Len = L - 1;
      long long val;
      if (L == 0)
        val = 1;
      else
        val = RSQ(1, n, 1);
      l_q = r_q = P[i].first;
      Len = L;
      X = val;
      Update(1, n, 1);
    }
  }
  cout << Seg[2][1] << endl;
  return 0;
}

```