#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> a;
vector<int> s;
const int roz = (1 << 21);
int get(int x) {
  int lo = 0;
  int hi = n - 1;
  while (lo + 1 < hi) {
    int mid = (lo + hi) / 2;
    if (s[mid] <= x)
      lo = mid;
    else
      hi = mid;
  }
  while (s[lo] != x) lo++;
  return lo;
}
void obrob() {
  s = a;
  sort(s.begin(), s.end());
  vector<int> res;
  for (int i = 0; i < n; i++) {
    int t = a[i];
    int pos = get(a[i]);
    res.push_back(pos + 1);
  }
  a = res;
}
int licz[roz];
void drzewoliczdodaj(int pos, int v) {
  pos += roz / 2;
  while (pos >= 1) {
    licz[pos] += v;
    pos /= 2;
  }
}
int getsumalicz(int lo, int hi, int u, int a, int b) {
  if (a == lo && b == hi) {
    return licz[u];
  }
  if (b <= a) return 0;
  int mid = (lo + hi) / 2;
  int x, y;
  x = getsumalicz(lo, mid, u * 2, a, min(b, mid));
  y = getsumalicz(mid, hi, u * 2 + 1, max(a, mid), b);
  return x + y;
}
long long int tree[roz];
void update(int lo, int hi, int u, int a, int b, int c) {
  if (a == lo && b == hi) {
    tree[u] += c;
    return;
  }
  if (b <= a) return;
  int mid = (lo + hi) / 2;
  update(lo, mid, u * 2, a, min(b, mid), c);
  update(mid, hi, u * 2 + 1, max(a, mid), b, c);
}
long long int suma(int v) {
  long long int ans = 0;
  v += roz / 2;
  while (v >= 1) {
    ans += tree[v];
    v /= 2;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(tree, 0, sizeof tree);
  memset(licz, 0, sizeof licz);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
  }
  obrob();
  long long int ans = 0;
  drzewoliczdodaj(a[0], 1);
  for (int i = 1; i < n; i++) {
    drzewoliczdodaj(a[i], 1);
    int t = getsumalicz(0, roz / 2, 1, a[i] + 1, n + 1);
    update(0, roz / 2, 1, 0, a[i], t);
    ans += suma(a[i]);
  }
  cout << ans;
  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;
int n;
vector<int> a;
vector<int> s;
const int roz = (1 << 21);
int get(int x) {
  int lo = 0;
  int hi = n - 1;
  while (lo + 1 < hi) {
    int mid = (lo + hi) / 2;
    if (s[mid] <= x)
      lo = mid;
    else
      hi = mid;
  }
  while (s[lo] != x) lo++;
  return lo;
}
void obrob() {
  s = a;
  sort(s.begin(), s.end());
  vector<int> res;
  for (int i = 0; i < n; i++) {
    int t = a[i];
    int pos = get(a[i]);
    res.push_back(pos + 1);
  }
  a = res;
}
int licz[roz];
void drzewoliczdodaj(int pos, int v) {
  pos += roz / 2;
  while (pos >= 1) {
    licz[pos] += v;
    pos /= 2;
  }
}
int getsumalicz(int lo, int hi, int u, int a, int b) {
  if (a == lo && b == hi) {
    return licz[u];
  }
  if (b <= a) return 0;
  int mid = (lo + hi) / 2;
  int x, y;
  x = getsumalicz(lo, mid, u * 2, a, min(b, mid));
  y = getsumalicz(mid, hi, u * 2 + 1, max(a, mid), b);
  return x + y;
}
long long int tree[roz];
void update(int lo, int hi, int u, int a, int b, int c) {
  if (a == lo && b == hi) {
    tree[u] += c;
    return;
  }
  if (b <= a) return;
  int mid = (lo + hi) / 2;
  update(lo, mid, u * 2, a, min(b, mid), c);
  update(mid, hi, u * 2 + 1, max(a, mid), b, c);
}
long long int suma(int v) {
  long long int ans = 0;
  v += roz / 2;
  while (v >= 1) {
    ans += tree[v];
    v /= 2;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(tree, 0, sizeof tree);
  memset(licz, 0, sizeof licz);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
  }
  obrob();
  long long int ans = 0;
  drzewoliczdodaj(a[0], 1);
  for (int i = 1; i < n; i++) {
    drzewoliczdodaj(a[i], 1);
    int t = getsumalicz(0, roz / 2, 1, a[i] + 1, n + 1);
    update(0, roz / 2, 1, 0, a[i], t);
    ans += suma(a[i]);
  }
  cout << ans;
  return 0;
}

```