#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
int fx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
int fy[] = {0, 0, -1, 1, -1, 1, -1, 1};
int kx[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky[] = {2, -2, 1, -1, 2, -2, 1, -1};
struct X {
  int val;
  int pos;
};
X A[1000006];
bool cmp(X a, X b) { return a.val < b.val; }
int tree[4 * 1000006];
void update(int node, int L, int R, int l, int r) {
  if (L > r || R < l) return;
  if (L == R && L == l) {
    tree[node] = 1;
    return;
  }
  int mid = (L + R) / 2;
  int left = 2 * node;
  int right = 2 * node + 1;
  update(left, L, mid, l, r);
  update(right, mid + 1, R, l, r);
  tree[node] = tree[left] + tree[right];
}
int query(int node, int L, int R, int l, int r) {
  if (L > r || R < l) return 0;
  if (L >= l && R <= r) return tree[node];
  int mid = (L + R) / 2;
  int left = 2 * node;
  int right = 2 * node + 1;
  return query(left, L, mid, l, r) + query(right, mid + 1, R, l, r);
}
int lmax[1000006], rmin[1000006];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i].val;
    A[i].pos = i;
  }
  sort(A + 1, A + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    int pos = A[i].pos;
    rmin[pos] = query(1, 1, n, pos + 1, n);
    update(1, 1, n, pos, pos);
  }
  for (int i = 1; i <= 4 * n; i++) tree[i] = 0;
  for (int i = n; i >= 1; i--) {
    int pos = A[i].pos;
    lmax[pos] = query(1, 1, n, 1, pos - 1);
    update(1, 1, n, pos, pos);
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) res += (long long)lmax[i] * (long long)rmin[i];
  cout << res << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 double eps = 1e-9;
int fx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
int fy[] = {0, 0, -1, 1, -1, 1, -1, 1};
int kx[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky[] = {2, -2, 1, -1, 2, -2, 1, -1};
struct X {
  int val;
  int pos;
};
X A[1000006];
bool cmp(X a, X b) { return a.val < b.val; }
int tree[4 * 1000006];
void update(int node, int L, int R, int l, int r) {
  if (L > r || R < l) return;
  if (L == R && L == l) {
    tree[node] = 1;
    return;
  }
  int mid = (L + R) / 2;
  int left = 2 * node;
  int right = 2 * node + 1;
  update(left, L, mid, l, r);
  update(right, mid + 1, R, l, r);
  tree[node] = tree[left] + tree[right];
}
int query(int node, int L, int R, int l, int r) {
  if (L > r || R < l) return 0;
  if (L >= l && R <= r) return tree[node];
  int mid = (L + R) / 2;
  int left = 2 * node;
  int right = 2 * node + 1;
  return query(left, L, mid, l, r) + query(right, mid + 1, R, l, r);
}
int lmax[1000006], rmin[1000006];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i].val;
    A[i].pos = i;
  }
  sort(A + 1, A + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    int pos = A[i].pos;
    rmin[pos] = query(1, 1, n, pos + 1, n);
    update(1, 1, n, pos, pos);
  }
  for (int i = 1; i <= 4 * n; i++) tree[i] = 0;
  for (int i = n; i >= 1; i--) {
    int pos = A[i].pos;
    lmax[pos] = query(1, 1, n, 1, pos - 1);
    update(1, 1, n, pos, pos);
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) res += (long long)lmax[i] * (long long)rmin[i];
  cout << res << "\n";
  return 0;
}

```