#include <bits/stdc++.h>
using namespace std;
map<int, int> q;
long long t[4000400], a[1000100], f[1000100];
long long ans = 0;
long long fsum(int l, int r, int lv, int rv, int v) {
  if (l > rv || r < lv) return 0;
  if (lv >= l && rv <= r) return t[v];
  return fsum(l, r, lv, (lv + rv) / 2, v * 2) +
         fsum(l, r, (lv + rv) / 2 + 1, rv, v * 2 + 1);
}
void modify(int v, long long d) {
  t[v] = d;
  v /= 2;
  for (; v > 0; v /= 2) t[v] = t[v * 2] + t[v * 2 + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) q[a[i]] = 0;
  int cnt = 0;
  for (map<int, int>::iterator it = q.begin(); it != q.end(); it++)
    it->second = ++cnt;
  for (int i = 0; i < n; i++) a[i] = q[a[i]];
  int z = 1;
  while (z < cnt) z *= 2;
  for (int i = 0; i < n; i++) {
    long long s = fsum(a[i] + 1, z, 1, z, 1);
    modify(a[i] + z - 1, 1);
    f[i] = s;
  }
  for (int i = 1; i < z * 2; i++) t[i] = 0;
  for (int i = 0; i < n; i++) {
    long long s = fsum(a[i] + 1, z, 1, z, 1);
    modify(a[i] + z - 1, f[i]);
    f[i] = s;
  }
  for (int i = 0; i < n; i++) ans += f[i];
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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;
map<int, int> q;
long long t[4000400], a[1000100], f[1000100];
long long ans = 0;
long long fsum(int l, int r, int lv, int rv, int v) {
  if (l > rv || r < lv) return 0;
  if (lv >= l && rv <= r) return t[v];
  return fsum(l, r, lv, (lv + rv) / 2, v * 2) +
         fsum(l, r, (lv + rv) / 2 + 1, rv, v * 2 + 1);
}
void modify(int v, long long d) {
  t[v] = d;
  v /= 2;
  for (; v > 0; v /= 2) t[v] = t[v * 2] + t[v * 2 + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) q[a[i]] = 0;
  int cnt = 0;
  for (map<int, int>::iterator it = q.begin(); it != q.end(); it++)
    it->second = ++cnt;
  for (int i = 0; i < n; i++) a[i] = q[a[i]];
  int z = 1;
  while (z < cnt) z *= 2;
  for (int i = 0; i < n; i++) {
    long long s = fsum(a[i] + 1, z, 1, z, 1);
    modify(a[i] + z - 1, 1);
    f[i] = s;
  }
  for (int i = 1; i < z * 2; i++) t[i] = 0;
  for (int i = 0; i < n; i++) {
    long long s = fsum(a[i] + 1, z, 1, z, 1);
    modify(a[i] + z - 1, f[i]);
    f[i] = s;
  }
  for (int i = 0; i < n; i++) ans += f[i];
  cout << ans << endl;
  return 0;
}

```