#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const long long MAX = 1e6 + 5;
const long long MAX2 = 11;
const int MOD = 1000000000 + 7;
const long long INF = 20000;
const int nr[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int nc[] = {0, 1, 0, -1, 1, -1, 1, -1};
int n, x[MAX], cnt, bt;
long long ans, le[MAX], ri[MAX], ret, let;
map<int, int> m;
set<int> s;
inline void updl(int i, int up) {
  for (; i <= n; i += (i & -i)) le[i] += up;
}
inline void updr(int i) {
  for (; i <= n; i += (i & -i)) ri[i]++;
}
inline void quel(int i) {
  let = 0;
  for (; i > 0; i -= (i & -i)) let += le[i];
}
inline void quer(int i) {
  ret = 0;
  for (; i > 0; i -= (i & -i)) ret += ri[i];
}
int main() {
  cout << fixed << setprecision(6);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> x[i], s.insert(x[i]);
  for (auto i : s) m[i] = ++cnt;
  for (int i = 1; i <= n; i++) x[i] = m[x[i]];
  for (int i = 2; i <= n; i++) updl(x[i], 1);
  quel(4);
  for (int i = 2; i <= n; i++) {
    updr(x[i - 1]);
    quer(x[i]);
    quel(x[i] - 1);
    ret = i - 1 - ret;
    ans += ret * let;
    updl(x[i], -1);
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize("O2")
using namespace std;
const long long MAX = 1e6 + 5;
const long long MAX2 = 11;
const int MOD = 1000000000 + 7;
const long long INF = 20000;
const int nr[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int nc[] = {0, 1, 0, -1, 1, -1, 1, -1};
int n, x[MAX], cnt, bt;
long long ans, le[MAX], ri[MAX], ret, let;
map<int, int> m;
set<int> s;
inline void updl(int i, int up) {
  for (; i <= n; i += (i & -i)) le[i] += up;
}
inline void updr(int i) {
  for (; i <= n; i += (i & -i)) ri[i]++;
}
inline void quel(int i) {
  let = 0;
  for (; i > 0; i -= (i & -i)) let += le[i];
}
inline void quer(int i) {
  ret = 0;
  for (; i > 0; i -= (i & -i)) ret += ri[i];
}
int main() {
  cout << fixed << setprecision(6);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> x[i], s.insert(x[i]);
  for (auto i : s) m[i] = ++cnt;
  for (int i = 1; i <= n; i++) x[i] = m[x[i]];
  for (int i = 2; i <= n; i++) updl(x[i], 1);
  quel(4);
  for (int i = 2; i <= n; i++) {
    updr(x[i - 1]);
    quer(x[i]);
    quel(x[i] - 1);
    ret = i - 1 - ret;
    ans += ret * let;
    updl(x[i], -1);
  }
  cout << ans << "\n";
  return 0;
}

```