#include <bits/stdc++.h>
using namespace std;
const long long mxN = 1e6 + 10, M = 1e9 + 7;
long long n, x;
long long v[mxN];
vector<long long> aux;
map<long long, long long> mp;
long long BIT1[mxN], BIT2[mxN];
void upd(long long BIT[], long long x, long long val) {
  while (x <= mxN) {
    BIT[x] += val;
    x += (x & -x);
  }
}
long long sum(long long BIT[], long long x) {
  long long ans = 0;
  while (x > 0) {
    ans += BIT[x];
    x -= (x & -x);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    v[i] = x;
    aux.push_back(x);
  }
  sort(aux.begin(), aux.end());
  long long idx = 0;
  for (long long i = 0; i < (long long)aux.size(); i++) {
    if (mp.find(aux[i]) == mp.end()) mp[aux[i]] = ++idx;
  }
  for (long long i = 0; i < n; i++) {
    v[i] = mp[v[i]];
  }
  long long ans = 0;
  for (long long i = n - 1; i >= 0; i--) {
    long long temp = sum(BIT2, v[i] - 1);
    ans += temp;
    temp = sum(BIT1, v[i] - 1);
    upd(BIT2, v[i], temp);
    upd(BIT1, v[i], 1);
  }
  cout << ans;
}

### 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 long long mxN = 1e6 + 10, M = 1e9 + 7;
long long n, x;
long long v[mxN];
vector<long long> aux;
map<long long, long long> mp;
long long BIT1[mxN], BIT2[mxN];
void upd(long long BIT[], long long x, long long val) {
  while (x <= mxN) {
    BIT[x] += val;
    x += (x & -x);
  }
}
long long sum(long long BIT[], long long x) {
  long long ans = 0;
  while (x > 0) {
    ans += BIT[x];
    x -= (x & -x);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    v[i] = x;
    aux.push_back(x);
  }
  sort(aux.begin(), aux.end());
  long long idx = 0;
  for (long long i = 0; i < (long long)aux.size(); i++) {
    if (mp.find(aux[i]) == mp.end()) mp[aux[i]] = ++idx;
  }
  for (long long i = 0; i < n; i++) {
    v[i] = mp[v[i]];
  }
  long long ans = 0;
  for (long long i = n - 1; i >= 0; i--) {
    long long temp = sum(BIT2, v[i] - 1);
    ans += temp;
    temp = sum(BIT1, v[i] - 1);
    upd(BIT2, v[i], temp);
    upd(BIT1, v[i], 1);
  }
  cout << ans;
}

```