#include <bits/stdc++.h>
using namespace std;
long long const MAX = 2147483647;
long long const REALMAX = 9223372036854775807;
vector<long long> SG;
void update(long long x) {
  while (x > 0) {
    SG[x]++;
    x /= 2;
  }
}
long long getsum(long long a, long long b, long long V, long long l,
                 long long r) {
  if (a > r || b < l) {
    return 0;
  }
  if (a <= l && r <= b) {
    return SG[V];
  }
  long long m = (l + r) / 2;
  long long left = getsum(a, b, 2 * V, l, m);
  long long right = getsum(a, b, 2 * V + 1, m + 1, r);
  return left + right;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a, b, sum = 0, count = 1;
  vector<long long> v, v1, L;
  map<long long, long long> vis;
  cin >> a;
  long long M = 1;
  while (M < a) {
    M *= 2;
  }
  SG.resize(M * 2, 0);
  for (long long i = (0); i < (a); i++) {
    cin >> b;
    v.push_back(b);
  }
  v1 = v;
  sort(v1.begin(), v1.end());
  for (long long i = (0); i < (a); i++) {
    if (!vis[v1[i]]) {
      vis[v1[i]] = count;
      count++;
    }
  }
  for (long long i = (0); i < (a); i++) {
    L.push_back(getsum(M + vis[v[i]] - 1, 2 * M - 1, 1, M, M * 2 - 1));
    update(M + vis[v[i]] - 1);
  }
  SG.clear();
  SG.resize(M * 2, 0);
  for (long long i = (a - 1); i >= 0; i--) {
    sum += getsum(M, M + vis[v[i]] - 1, 1, M, M * 2 - 1) * L[i];
    update(M + vis[v[i]] - 1);
  }
  cout << sum << endl;
}

### Prompt

Please provide a CPP coded solution 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 const MAX = 2147483647;
long long const REALMAX = 9223372036854775807;
vector<long long> SG;
void update(long long x) {
  while (x > 0) {
    SG[x]++;
    x /= 2;
  }
}
long long getsum(long long a, long long b, long long V, long long l,
                 long long r) {
  if (a > r || b < l) {
    return 0;
  }
  if (a <= l && r <= b) {
    return SG[V];
  }
  long long m = (l + r) / 2;
  long long left = getsum(a, b, 2 * V, l, m);
  long long right = getsum(a, b, 2 * V + 1, m + 1, r);
  return left + right;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a, b, sum = 0, count = 1;
  vector<long long> v, v1, L;
  map<long long, long long> vis;
  cin >> a;
  long long M = 1;
  while (M < a) {
    M *= 2;
  }
  SG.resize(M * 2, 0);
  for (long long i = (0); i < (a); i++) {
    cin >> b;
    v.push_back(b);
  }
  v1 = v;
  sort(v1.begin(), v1.end());
  for (long long i = (0); i < (a); i++) {
    if (!vis[v1[i]]) {
      vis[v1[i]] = count;
      count++;
    }
  }
  for (long long i = (0); i < (a); i++) {
    L.push_back(getsum(M + vis[v[i]] - 1, 2 * M - 1, 1, M, M * 2 - 1));
    update(M + vis[v[i]] - 1);
  }
  SG.clear();
  SG.resize(M * 2, 0);
  for (long long i = (a - 1); i >= 0; i--) {
    sum += getsum(M, M + vis[v[i]] - 1, 1, M, M * 2 - 1) * L[i];
    update(M + vis[v[i]] - 1);
  }
  cout << sum << endl;
}

```