#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int N = 1000002;
const double PI = 4 * atan(1);
const double eps = 1e-7;
const long long oo = 1e18;
long long n, m;
long long tree[N][2];
long long a[N];
set<long long> ss;
map<long long, long long> maa, mal;
long long cnt;
long long get(long long u, long long ty) {
  long long to = 0;
  for (; u > 0; u -= (u) & (-u)) {
    to += tree[u][ty];
  }
  return to;
}
void update(long long u, long long ty, long long val) {
  for (; u <= cnt; u += (u) & (-u)) {
    tree[u][ty] += val;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ss.insert(a[i]);
  }
  cnt = 0;
  for (auto t : ss) {
    maa[t] = cnt;
    mal[cnt] = t;
    cnt++;
  }
  for (int i = 0; i < n; i++) {
    a[i] = cnt - maa[a[i]];
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    update(a[i], 0, 1);
    long long x = get(a[i] - 1, 0);
    update(a[i], 1, x);
    x = get(a[i] - 1, 1);
    ans += x;
  }
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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;
const int MOD = 1000000007;
const int N = 1000002;
const double PI = 4 * atan(1);
const double eps = 1e-7;
const long long oo = 1e18;
long long n, m;
long long tree[N][2];
long long a[N];
set<long long> ss;
map<long long, long long> maa, mal;
long long cnt;
long long get(long long u, long long ty) {
  long long to = 0;
  for (; u > 0; u -= (u) & (-u)) {
    to += tree[u][ty];
  }
  return to;
}
void update(long long u, long long ty, long long val) {
  for (; u <= cnt; u += (u) & (-u)) {
    tree[u][ty] += val;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ss.insert(a[i]);
  }
  cnt = 0;
  for (auto t : ss) {
    maa[t] = cnt;
    mal[cnt] = t;
    cnt++;
  }
  for (int i = 0; i < n; i++) {
    a[i] = cnt - maa[a[i]];
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    update(a[i], 0, 1);
    long long x = get(a[i] - 1, 0);
    update(a[i], 1, x);
    x = get(a[i] - 1, 1);
    ans += x;
  }
  cout << ans;
  return 0;
}

```