#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int N = 1e6;
struct Segtree {
  int n;
  vector<long long> st;
  const long long default_value = 0;
  long long op(long long l, long long r) { return l + r; }
  void build() {
    for (int i = n - 1; i > 0; --i) st[i] = op(st[i << 1], st[i << 1 | 1]);
  }
  void modify(int p, long long value) {
    for (st[p += n] = value; p > 1; p >>= 1) st[p >> 1] = op(st[p], st[p ^ 1]);
  }
  long long query(int l, int r) {
    long long res = default_value;
    for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
      if (l & 1) res = op(st[l++], res);
      if (r & 1) res = op(res, st[--r]);
    }
    return res;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n, 0);
  for (int i_ = 0; i_ < (int)a.size(); ++i_) cin >> a[i_];
  Segtree T[2];
  T[0].n = n, T[1].n = n;
  T[0].st.assign(2 * N, 0);
  T[1].st.assign(2 * N, 0);
  vector<int> I(n);
  for (int i = 0; i < n; ++i) {
    I[i] = i;
    T[0].st[n + i] = 0;
    T[1].st[n + i] = 0;
  }
  T[0].build(), T[1].build();
  sort(I.begin(), I.end(), [&a](int i, int j) { return a[i] > a[j]; });
  long long ans = 0;
  for (auto i : I) {
    ans += T[1].query(0, i);
    long long q = T[0].query(0, i);
    T[0].modify(i, 1);
    T[1].modify(i, q);
  }
  cout << ans << '\n';
  exit(0);
}

### Prompt

Please formulate 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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int N = 1e6;
struct Segtree {
  int n;
  vector<long long> st;
  const long long default_value = 0;
  long long op(long long l, long long r) { return l + r; }
  void build() {
    for (int i = n - 1; i > 0; --i) st[i] = op(st[i << 1], st[i << 1 | 1]);
  }
  void modify(int p, long long value) {
    for (st[p += n] = value; p > 1; p >>= 1) st[p >> 1] = op(st[p], st[p ^ 1]);
  }
  long long query(int l, int r) {
    long long res = default_value;
    for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
      if (l & 1) res = op(st[l++], res);
      if (r & 1) res = op(res, st[--r]);
    }
    return res;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n, 0);
  for (int i_ = 0; i_ < (int)a.size(); ++i_) cin >> a[i_];
  Segtree T[2];
  T[0].n = n, T[1].n = n;
  T[0].st.assign(2 * N, 0);
  T[1].st.assign(2 * N, 0);
  vector<int> I(n);
  for (int i = 0; i < n; ++i) {
    I[i] = i;
    T[0].st[n + i] = 0;
    T[1].st[n + i] = 0;
  }
  T[0].build(), T[1].build();
  sort(I.begin(), I.end(), [&a](int i, int j) { return a[i] > a[j]; });
  long long ans = 0;
  for (auto i : I) {
    ans += T[1].query(0, i);
    long long q = T[0].query(0, i);
    T[0].modify(i, 1);
    T[1].modify(i, q);
  }
  cout << ans << '\n';
  exit(0);
}

```