#include <bits/stdc++.h>
using namespace std;
const int N = 1000 * 1000 + 10;
int n, a[N];
long long ans;
struct node {
  int cnt;
  long long cnt_pr;
  node() { cnt = cnt_pr = 0; }
} seg[N << 2];
void upd(int x, int l = 0, int r = n, int id = 1) {
  int mid = l + r >> 1;
  if (l > x || r <= x) return;
  if (r - l == 1) return seg[id].cnt = 1, void();
  upd(x, l, mid, id << 1), upd(x, mid, r, id << 1 | 1);
  seg[id].cnt = seg[id << 1].cnt + seg[id << 1 | 1].cnt;
}
void upd_pr(int x, long long res = 0, int l = 0, int r = n, int id = 1) {
  int mid = l + r >> 1;
  if (r - l == 1) return seg[id].cnt_pr = res, void();
  if (x < mid)
    upd_pr(x, res + seg[id << 1 | 1].cnt, l, mid, id << 1);
  else
    upd_pr(x, res, mid, r, id << 1 | 1);
  seg[id].cnt_pr = seg[id << 1].cnt_pr + seg[id << 1 | 1].cnt_pr;
}
long long get(int x, long long res = 0, int l = 0, int r = n, int id = 1) {
  int mid = l + r >> 1;
  if (r - l == 1) return res;
  if (x < mid)
    return get(x, res + seg[id << 1 | 1].cnt_pr, l, mid, id << 1);
  else
    return get(x, res, mid, r, id << 1 | 1);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n;
  vector<int> vec(n);
  for (int i = 0; i < n; i++) cin >> a[i], vec[i] = a[i];
  sort(vec.begin(), vec.end());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(vec.begin(), vec.end(), a[i]) - vec.begin();
  for (int i = 0; i < n; i++) {
    upd(a[i]), upd_pr(a[i]);
    ans += get(a[i]);
  }
  cout << ans << '\n';
}

### 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;
const int N = 1000 * 1000 + 10;
int n, a[N];
long long ans;
struct node {
  int cnt;
  long long cnt_pr;
  node() { cnt = cnt_pr = 0; }
} seg[N << 2];
void upd(int x, int l = 0, int r = n, int id = 1) {
  int mid = l + r >> 1;
  if (l > x || r <= x) return;
  if (r - l == 1) return seg[id].cnt = 1, void();
  upd(x, l, mid, id << 1), upd(x, mid, r, id << 1 | 1);
  seg[id].cnt = seg[id << 1].cnt + seg[id << 1 | 1].cnt;
}
void upd_pr(int x, long long res = 0, int l = 0, int r = n, int id = 1) {
  int mid = l + r >> 1;
  if (r - l == 1) return seg[id].cnt_pr = res, void();
  if (x < mid)
    upd_pr(x, res + seg[id << 1 | 1].cnt, l, mid, id << 1);
  else
    upd_pr(x, res, mid, r, id << 1 | 1);
  seg[id].cnt_pr = seg[id << 1].cnt_pr + seg[id << 1 | 1].cnt_pr;
}
long long get(int x, long long res = 0, int l = 0, int r = n, int id = 1) {
  int mid = l + r >> 1;
  if (r - l == 1) return res;
  if (x < mid)
    return get(x, res + seg[id << 1 | 1].cnt_pr, l, mid, id << 1);
  else
    return get(x, res, mid, r, id << 1 | 1);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n;
  vector<int> vec(n);
  for (int i = 0; i < n; i++) cin >> a[i], vec[i] = a[i];
  sort(vec.begin(), vec.end());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(vec.begin(), vec.end(), a[i]) - vec.begin();
  for (int i = 0; i < n; i++) {
    upd(a[i]), upd_pr(a[i]);
    ans += get(a[i]);
  }
  cout << ans << '\n';
}

```