#include <bits/stdc++.h>
using namespace std;
void files(bool in, char *s) {
  if (in)
    freopen(s, "r", stdin);
  else
    freopen(s, "w", stdout);
}
const double PI = 4 * atan(1.0);
const double EPS = 1e-9;
const int infi = (int)1e9;
const long long infl = (long long)1e17;
const int MOD = (int)1e9 + 7;
const int MAX = (int)1e6 + 22;
const int dx[] = {0, 1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 0, -1, 1, 1, -1, 1, -1};
const int kdx[] = {1, 1, -1, -1, 2, 2, -2, -2};
const int kdy[] = {2, -2, 2, -2, 1, -1, 1, -1};
long long n;
long long a[MAX];
pair<long long, long long> p[MAX];
long long seg[4 * MAX];
void update(long long st, long long en, long long idx, long long l,
            long long r) {
  if (l > en || r < st) return;
  if (l <= st && en <= r) {
    seg[idx]++;
    return;
  }
  long long mid = (st + en) / 2;
  update(st, mid, 2 * idx, l, r);
  update(mid + 1, en, 2 * idx + 1, l, r);
}
long long query(long long st, long long en, long long idx, long long x) {
  if (st != en && seg[idx]) {
    seg[2 * idx] += seg[idx];
    seg[2 * idx + 1] += seg[idx];
    seg[idx] = 0;
  }
  if (x > en || x < st) return 0;
  if (st == en) return (st == x ? seg[idx] : 0);
  long long mid = (st + en) / 2;
  long long lf = query(st, mid, 2 * idx, x);
  long long rt = query(mid + 1, en, 2 * idx + 1, x);
  return lf + rt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].first;
    p[i].second = i;
  }
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; i++) a[p[i].second] = i;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long bfr_grt = query(1, n, 1, a[i]);
    long long bfr_lss = i - bfr_grt - 1;
    long long atr_lss = (a[i] - 1 - bfr_lss);
    ans += bfr_grt * atr_lss;
    update(1, n, 1, 1, a[i]);
  }
  cout << ans << "\n";
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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;
void files(bool in, char *s) {
  if (in)
    freopen(s, "r", stdin);
  else
    freopen(s, "w", stdout);
}
const double PI = 4 * atan(1.0);
const double EPS = 1e-9;
const int infi = (int)1e9;
const long long infl = (long long)1e17;
const int MOD = (int)1e9 + 7;
const int MAX = (int)1e6 + 22;
const int dx[] = {0, 1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 0, -1, 1, 1, -1, 1, -1};
const int kdx[] = {1, 1, -1, -1, 2, 2, -2, -2};
const int kdy[] = {2, -2, 2, -2, 1, -1, 1, -1};
long long n;
long long a[MAX];
pair<long long, long long> p[MAX];
long long seg[4 * MAX];
void update(long long st, long long en, long long idx, long long l,
            long long r) {
  if (l > en || r < st) return;
  if (l <= st && en <= r) {
    seg[idx]++;
    return;
  }
  long long mid = (st + en) / 2;
  update(st, mid, 2 * idx, l, r);
  update(mid + 1, en, 2 * idx + 1, l, r);
}
long long query(long long st, long long en, long long idx, long long x) {
  if (st != en && seg[idx]) {
    seg[2 * idx] += seg[idx];
    seg[2 * idx + 1] += seg[idx];
    seg[idx] = 0;
  }
  if (x > en || x < st) return 0;
  if (st == en) return (st == x ? seg[idx] : 0);
  long long mid = (st + en) / 2;
  long long lf = query(st, mid, 2 * idx, x);
  long long rt = query(mid + 1, en, 2 * idx + 1, x);
  return lf + rt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].first;
    p[i].second = i;
  }
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; i++) a[p[i].second] = i;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long bfr_grt = query(1, n, 1, a[i]);
    long long bfr_lss = i - bfr_grt - 1;
    long long atr_lss = (a[i] - 1 - bfr_lss);
    ans += bfr_grt * atr_lss;
    update(1, n, 1, 1, a[i]);
  }
  cout << ans << "\n";
}

```