#include <bits/stdc++.h>
using namespace std;
long long n, a[1000006], a1[1000006];
long long tree[4000006], lo[1000006], hi[1000006];
void upd(int x, int l, int r, int p) {
  if (l > r || p < l || p > r) return;
  if (l == p && r == p) {
    ++tree[x];
    return;
  }
  upd((x << 1), l, ((l + r) >> 1), p);
  upd((x << 1) + 1, ((l + r) >> 1) + 1, r, p);
  tree[x] = tree[x << 1] + tree[(x << 1) + 1];
}
void upd(int p) { upd(1, 0, n + 1, p); }
long long query(int x, int l, int r, int p, int q) {
  if (l > r || q < l || p > r) return 0LL;
  if (p <= l && r <= q) return tree[x];
  return query((x << 1), l, ((l + r) >> 1), p, q) +
         query((x << 1) + 1, ((l + r) >> 1) + 1, r, p, q);
}
long long query(int p, int q) { return query(1, 0, n + 1, p, q); }
int main() {
  long long ans = 0;
  scanf("%lld", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    a1[i] = a[i];
  }
  sort(a1 + 1, a1 + n + 1);
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(a1 + 1, a1 + n + 1, a[i]) - a1;
  }
  for (int i = 1; i <= n; ++i) {
    hi[i] = query(a[i] + 1, n + 1);
    upd(a[i]);
  }
  memset(tree, 0, sizeof(tree));
  for (int i = n; i > 0; --i) {
    lo[i] = query(0, a[i] - 1);
    upd(a[i]);
  }
  for (int i = 1; i <= n; ++i) {
    ans += lo[i] * hi[i];
  }
  printf("%lld", ans);
  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 n, a[1000006], a1[1000006];
long long tree[4000006], lo[1000006], hi[1000006];
void upd(int x, int l, int r, int p) {
  if (l > r || p < l || p > r) return;
  if (l == p && r == p) {
    ++tree[x];
    return;
  }
  upd((x << 1), l, ((l + r) >> 1), p);
  upd((x << 1) + 1, ((l + r) >> 1) + 1, r, p);
  tree[x] = tree[x << 1] + tree[(x << 1) + 1];
}
void upd(int p) { upd(1, 0, n + 1, p); }
long long query(int x, int l, int r, int p, int q) {
  if (l > r || q < l || p > r) return 0LL;
  if (p <= l && r <= q) return tree[x];
  return query((x << 1), l, ((l + r) >> 1), p, q) +
         query((x << 1) + 1, ((l + r) >> 1) + 1, r, p, q);
}
long long query(int p, int q) { return query(1, 0, n + 1, p, q); }
int main() {
  long long ans = 0;
  scanf("%lld", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    a1[i] = a[i];
  }
  sort(a1 + 1, a1 + n + 1);
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(a1 + 1, a1 + n + 1, a[i]) - a1;
  }
  for (int i = 1; i <= n; ++i) {
    hi[i] = query(a[i] + 1, n + 1);
    upd(a[i]);
  }
  memset(tree, 0, sizeof(tree));
  for (int i = n; i > 0; --i) {
    lo[i] = query(0, a[i] - 1);
    upd(a[i]);
  }
  for (int i = 1; i <= n; ++i) {
    ans += lo[i] * hi[i];
  }
  printf("%lld", ans);
  return 0;
}

```