#include <bits/stdc++.h>
using namespace std;
pair<int, int> p[1000001];
int n, High, a[1000001], SegTree[8000001], fix[1000001];
long long ans, Left[1000001], Right[1000001];
void upd(int ind, int low, int high, int pos) {
  int mid = (low + high) / 2;
  if (low > high) return;
  if (ind < low || ind > high) return;
  if (ind <= low && ind >= high) {
    SegTree[pos] = fix[ind];
    return;
  }
  upd(ind, low, mid, 2 * pos + 1);
  upd(ind, mid + 1, high, 2 * pos + 2);
  SegTree[pos] = SegTree[2 * pos + 1] + SegTree[2 * pos + 2];
}
int sum(int L, int R, int low, int high, int pos) {
  int mid = (low + high) / 2;
  if (low > high) return 0;
  if (L > high || R < low) return 0;
  if (L <= low && R >= high) {
    return SegTree[pos];
  }
  return sum(L, R, low, mid, 2 * pos + 1) +
         sum(L, R, mid + 1, high, 2 * pos + 2);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], p[i].first = a[i], p[i].second = i;
  sort(p, p + n);
  for (int i = 0; i < n; i++) {
    a[p[i].second] = i + 1;
  }
  for (int i = 0; i < n; i++) {
    int x = log2(a[i]);
    if ((1 << x) == a[i])
      High = max(High, a[i] - 1);
    else
      High = max(High, (1 << x + 1) - 1);
  }
  for (int i = 0; i < n; i++) {
    fix[a[i] - 1] = 1;
    upd(a[i] - 1, 0, High, 0);
    Left[i] = sum(a[i], High, 0, High, 0);
  }
  for (int i = 0; i < n; i++) fix[i] = 0;
  for (int i = 0; i < 3 * n; i++) SegTree[i] = 0;
  for (int i = n - 1; i > -1; i--) {
    fix[a[i] - 1] = 1;
    int x = log2(a[i]);
    upd(a[i] - 1, 0, High, 0);
    Right[i] = sum(0, a[i] - 2, 0, High, 0);
  }
  for (int i = 0; i < n; i++) {
    ans += Left[i] * Right[i];
  }
  cout << ans;
}

### 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;
pair<int, int> p[1000001];
int n, High, a[1000001], SegTree[8000001], fix[1000001];
long long ans, Left[1000001], Right[1000001];
void upd(int ind, int low, int high, int pos) {
  int mid = (low + high) / 2;
  if (low > high) return;
  if (ind < low || ind > high) return;
  if (ind <= low && ind >= high) {
    SegTree[pos] = fix[ind];
    return;
  }
  upd(ind, low, mid, 2 * pos + 1);
  upd(ind, mid + 1, high, 2 * pos + 2);
  SegTree[pos] = SegTree[2 * pos + 1] + SegTree[2 * pos + 2];
}
int sum(int L, int R, int low, int high, int pos) {
  int mid = (low + high) / 2;
  if (low > high) return 0;
  if (L > high || R < low) return 0;
  if (L <= low && R >= high) {
    return SegTree[pos];
  }
  return sum(L, R, low, mid, 2 * pos + 1) +
         sum(L, R, mid + 1, high, 2 * pos + 2);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], p[i].first = a[i], p[i].second = i;
  sort(p, p + n);
  for (int i = 0; i < n; i++) {
    a[p[i].second] = i + 1;
  }
  for (int i = 0; i < n; i++) {
    int x = log2(a[i]);
    if ((1 << x) == a[i])
      High = max(High, a[i] - 1);
    else
      High = max(High, (1 << x + 1) - 1);
  }
  for (int i = 0; i < n; i++) {
    fix[a[i] - 1] = 1;
    upd(a[i] - 1, 0, High, 0);
    Left[i] = sum(a[i], High, 0, High, 0);
  }
  for (int i = 0; i < n; i++) fix[i] = 0;
  for (int i = 0; i < 3 * n; i++) SegTree[i] = 0;
  for (int i = n - 1; i > -1; i--) {
    fix[a[i] - 1] = 1;
    int x = log2(a[i]);
    upd(a[i] - 1, 0, High, 0);
    Right[i] = sum(0, a[i] - 2, 0, High, 0);
  }
  for (int i = 0; i < n; i++) {
    ans += Left[i] * Right[i];
  }
  cout << ans;
}

```