#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const int N = 1e6 + 6;
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b > 0) {
    long long t = a % b;
    a = b, b = t;
  }
  return a;
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long is_prime(long long n) {
  if (n <= 1 || n > 3 && (n % 2 == 0 || n % 3 == 0)) return 0;
  for (long long i = 5, t = 2; i * i <= n; i += t, t = 6 - t)
    if (n % i == 0) return 0;
  return 1;
}
long long bit[5000005];
long long N1 = 5000004;
void update(int x, int del) {
  for (; x <= N1; x += x & -x) {
    bit[x] += del;
  }
}
int query(int x) {
  int sum = 0;
  for (; x > 0; x -= x & -x) {
    sum += bit[x];
  }
  return sum;
}
int main() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> v;
  long long x;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    v.push_back(make_pair(x, i + 1));
  }
  vector<long long> p;
  long long g[n + 1], h[n + 1];
  memset(g, 0, sizeof(g));
  memset(h, 0, sizeof(h));
  sort((v).begin(), (v).end());
  for (long long i = 0; i < n; i++) {
    p.push_back(v[i].second);
  }
  for (int i = 0; i < n; i++) {
    g[i + 1] = i - query(p[i]);
    update(p[i], 1);
  }
  memset(bit, 0, sizeof(bit));
  for (int i = n - 1; i >= 0; i--) {
    h[i + 1] = query(p[i]);
    update(p[i], 1);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += (g[i] * (h[i]));
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

In CPP, your task is to solve 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 = 1e9 + 7;
const int INF = 1e9;
const int N = 1e6 + 6;
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b > 0) {
    long long t = a % b;
    a = b, b = t;
  }
  return a;
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long is_prime(long long n) {
  if (n <= 1 || n > 3 && (n % 2 == 0 || n % 3 == 0)) return 0;
  for (long long i = 5, t = 2; i * i <= n; i += t, t = 6 - t)
    if (n % i == 0) return 0;
  return 1;
}
long long bit[5000005];
long long N1 = 5000004;
void update(int x, int del) {
  for (; x <= N1; x += x & -x) {
    bit[x] += del;
  }
}
int query(int x) {
  int sum = 0;
  for (; x > 0; x -= x & -x) {
    sum += bit[x];
  }
  return sum;
}
int main() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> v;
  long long x;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    v.push_back(make_pair(x, i + 1));
  }
  vector<long long> p;
  long long g[n + 1], h[n + 1];
  memset(g, 0, sizeof(g));
  memset(h, 0, sizeof(h));
  sort((v).begin(), (v).end());
  for (long long i = 0; i < n; i++) {
    p.push_back(v[i].second);
  }
  for (int i = 0; i < n; i++) {
    g[i + 1] = i - query(p[i]);
    update(p[i], 1);
  }
  memset(bit, 0, sizeof(bit));
  for (int i = n - 1; i >= 0; i--) {
    h[i + 1] = query(p[i]);
    update(p[i], 1);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += (g[i] * (h[i]));
  }
  cout << ans << "\n";
  return 0;
}

```