#include <bits/stdc++.h>
const long long md = 1e9 + 7;
const int Inf = 1e9;
const long long Inf64 = 1e18;
const long long MaxN = 1e6 + 100;
const long double eps = 1e-15;
using namespace std;
long long Tree[4 * MaxN];
long long getSum(long long l, long long r, long long x, long long y,
                 long long id) {
  if (l == x && r == y) {
    return Tree[id];
  } else {
    long long mid = (l + r) / 2;
    if (mid >= y) {
      return getSum(l, mid, x, y, 2 * id + 1);
    } else if (mid + 1 <= x) {
      return getSum(mid + 1, r, x, y, 2 * id + 2);
    } else {
      return getSum(l, mid, x, mid, 2 * id + 1) +
             getSum(mid + 1, r, mid + 1, y, 2 * id + 2);
    }
  }
}
void edit(long long l, long long r, long long x, long long id) {
  if (l == r) {
    Tree[id]++;
  } else {
    long long mid = (l + r) / 2;
    if (mid >= x) {
      edit(l, mid, x, 2 * id + 1);
      Tree[id] = Tree[2 * id + 1] + Tree[2 * id + 2];
    } else {
      edit(mid + 1, r, x, 2 * id + 2);
      Tree[id] = Tree[2 * id + 1] + Tree[2 * id + 2];
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long N;
  cin >> N;
  vector<pair<long long, long long> > Zip(N);
  for (int i = 0; i < N; i++) {
    long long a;
    cin >> a;
    Zip[i] = {a, i};
  }
  sort(Zip.begin(), Zip.end());
  vector<long long> Ans(N);
  for (int i = 0; i < N; i++) {
    long long t = getSum(0, N - 1, Zip[i].second, N - 1, 0);
    Ans[Zip[i].second] = t;
    edit(0, N - 1, Zip[i].second, 0);
  }
  fill(Tree, Tree + 4 * MaxN, 0);
  reverse(Zip.begin(), Zip.end());
  long long Res = 0;
  for (int i = 0; i < N; i++) {
    long long t = getSum(0, N - 1, 0, Zip[i].second, 0);
    Res += Ans[Zip[i].second] * t;
    edit(0, N - 1, Zip[i].second, 0);
  }
  cout << Res;
  cerr << '\n'
       << "Time execute: " << clock() / (double)CLOCKS_PER_SEC << " sec"
       << '\n';
  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>
const long long md = 1e9 + 7;
const int Inf = 1e9;
const long long Inf64 = 1e18;
const long long MaxN = 1e6 + 100;
const long double eps = 1e-15;
using namespace std;
long long Tree[4 * MaxN];
long long getSum(long long l, long long r, long long x, long long y,
                 long long id) {
  if (l == x && r == y) {
    return Tree[id];
  } else {
    long long mid = (l + r) / 2;
    if (mid >= y) {
      return getSum(l, mid, x, y, 2 * id + 1);
    } else if (mid + 1 <= x) {
      return getSum(mid + 1, r, x, y, 2 * id + 2);
    } else {
      return getSum(l, mid, x, mid, 2 * id + 1) +
             getSum(mid + 1, r, mid + 1, y, 2 * id + 2);
    }
  }
}
void edit(long long l, long long r, long long x, long long id) {
  if (l == r) {
    Tree[id]++;
  } else {
    long long mid = (l + r) / 2;
    if (mid >= x) {
      edit(l, mid, x, 2 * id + 1);
      Tree[id] = Tree[2 * id + 1] + Tree[2 * id + 2];
    } else {
      edit(mid + 1, r, x, 2 * id + 2);
      Tree[id] = Tree[2 * id + 1] + Tree[2 * id + 2];
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long N;
  cin >> N;
  vector<pair<long long, long long> > Zip(N);
  for (int i = 0; i < N; i++) {
    long long a;
    cin >> a;
    Zip[i] = {a, i};
  }
  sort(Zip.begin(), Zip.end());
  vector<long long> Ans(N);
  for (int i = 0; i < N; i++) {
    long long t = getSum(0, N - 1, Zip[i].second, N - 1, 0);
    Ans[Zip[i].second] = t;
    edit(0, N - 1, Zip[i].second, 0);
  }
  fill(Tree, Tree + 4 * MaxN, 0);
  reverse(Zip.begin(), Zip.end());
  long long Res = 0;
  for (int i = 0; i < N; i++) {
    long long t = getSum(0, N - 1, 0, Zip[i].second, 0);
    Res += Ans[Zip[i].second] * t;
    edit(0, N - 1, Zip[i].second, 0);
  }
  cout << Res;
  cerr << '\n'
       << "Time execute: " << clock() / (double)CLOCKS_PER_SEC << " sec"
       << '\n';
  return 0;
}

```