#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
const long long mod = 998244353;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long binpow(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
const int maxn = 1e6 + 5;
int bit[maxn], pref[maxn], suf[maxn];
int sum(int idx) {
  int ret = 0;
  for (++idx; idx > 0; idx -= idx & -idx) ret += bit[idx];
  return ret;
}
int sum(int l, int r) { return sum(r) - sum(l - 1); }
void add(int idx, int delta) {
  for (++idx; idx < maxn; idx += idx & -idx) bit[idx] += delta;
}
void solve() {
  int n;
  cin >> n;
  vector<int> v(n), st;
  {
    for (auto &_re : v) cin >> _re;
  };
  st = v;
  sort(st.begin(), st.end());
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    m[st[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    v[i] = m[v[i]];
    pref[i] = sum(v[i] + 1, n + 1);
    add(v[i], 1);
  }
  memset(bit, 0, sizeof(bit));
  for (int i = n - 1; i >= 0; i--) {
    suf[i] = sum(v[i] - 1);
    add(v[i], 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += (long long)pref[i] * suf[i];
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

### 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 long long M = 1e9 + 7;
const long long mod = 998244353;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long binpow(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
const int maxn = 1e6 + 5;
int bit[maxn], pref[maxn], suf[maxn];
int sum(int idx) {
  int ret = 0;
  for (++idx; idx > 0; idx -= idx & -idx) ret += bit[idx];
  return ret;
}
int sum(int l, int r) { return sum(r) - sum(l - 1); }
void add(int idx, int delta) {
  for (++idx; idx < maxn; idx += idx & -idx) bit[idx] += delta;
}
void solve() {
  int n;
  cin >> n;
  vector<int> v(n), st;
  {
    for (auto &_re : v) cin >> _re;
  };
  st = v;
  sort(st.begin(), st.end());
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    m[st[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    v[i] = m[v[i]];
    pref[i] = sum(v[i] + 1, n + 1);
    add(v[i], 1);
  }
  memset(bit, 0, sizeof(bit));
  for (int i = n - 1; i >= 0; i--) {
    suf[i] = sum(v[i] - 1);
    add(v[i], 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += (long long)pref[i] * suf[i];
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

```