#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)1e6 + 1000;
long long t[maxn];
void upd(int pos, long long val) {
  for (; pos < maxn; pos |= pos + 1) t[pos] += val;
}
long long get(int l, int r) {
  long long res = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) res += t[r];
  l--;
  for (; l >= 0; l = (l & (l + 1)) - 1) res -= t[l];
  return res;
}
int a[maxn], n, tst;
vector<int> srt;
long long G[maxn];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), srt.push_back(a[i]);
  sort(srt.begin(), srt.end());
  srt.resize(unique(srt.begin(), srt.end()) - srt.begin());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(srt.begin(), srt.end(), a[i]) - srt.begin() + 1;
  long long ans = 0;
  for (int i = 1; i <= n; i++) G[i] = get(a[i] + 1, maxn - 1), upd(a[i], 1);
  memset(t, 0, sizeof(t));
  for (int i = 1; i <= n; i++) ans += get(a[i] + 1, maxn - 1), upd(a[i], G[i]);
  printf("%lld\n", ans);
  return 0;
}

### 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 int maxn = (int)1e6 + 1000;
long long t[maxn];
void upd(int pos, long long val) {
  for (; pos < maxn; pos |= pos + 1) t[pos] += val;
}
long long get(int l, int r) {
  long long res = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) res += t[r];
  l--;
  for (; l >= 0; l = (l & (l + 1)) - 1) res -= t[l];
  return res;
}
int a[maxn], n, tst;
vector<int> srt;
long long G[maxn];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), srt.push_back(a[i]);
  sort(srt.begin(), srt.end());
  srt.resize(unique(srt.begin(), srt.end()) - srt.begin());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(srt.begin(), srt.end(), a[i]) - srt.begin() + 1;
  long long ans = 0;
  for (int i = 1; i <= n; i++) G[i] = get(a[i] + 1, maxn - 1), upd(a[i], 1);
  memset(t, 0, sizeof(t));
  for (int i = 1; i <= n; i++) ans += get(a[i] + 1, maxn - 1), upd(a[i], G[i]);
  printf("%lld\n", ans);
  return 0;
}

```