#include <bits/stdc++.h>
using namespace std;
vector<int> a(1000001);
vector<vector<int> > seg(4 * 1000001);
vector<int> merg(int node) {
  int l = 2 * node, r = 2 * node + 1;
  int sz1 = seg[l].size();
  int sz2 = seg[r].size();
  int k = 0, i = 0, j = 0;
  while (i < sz1 && j < sz2) {
    if (seg[l][i] < seg[r][j])
      seg[node].push_back(seg[l][i++]);
    else
      seg[node].push_back(seg[r][j++]);
  }
  if (i < sz1) {
    while (i < sz1) {
      seg[node].push_back(seg[l][i++]);
    }
  }
  if (j < sz2) {
    while (j < sz2) {
      seg[node].push_back(seg[r][j++]);
    }
  }
  return seg[node];
}
void build(int cur, int st, int end) {
  if (st == end) {
    vector<int> temp;
    temp.push_back(a[st]);
    seg[cur] = temp;
    return;
  }
  long long mid = (st + end) >> 1;
  build(2 * cur, st, mid);
  build(2 * cur + 1, mid + 1, end);
  merg(cur);
}
int query2(int cur, int st, int end, int l, int r, int k) {
  if (l <= st && r >= end) {
    int temp =
        lower_bound(seg[cur].begin(), seg[cur].end(), k) - seg[cur].begin();
    return seg[cur].size() - temp;
  }
  if (r < st || l > end) return 0;
  long long mid = (st + end) >> 1;
  long long ans1 = query2(2 * cur, st, mid, l, r, k);
  long long ans2 = query2(2 * cur + 1, mid + 1, end, l, r, k);
  return (ans1 + ans2);
}
int main() {
  iostream::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (long long i = (0); i < (n); ++i) cin >> a[i];
  build(1, 0, n - 1);
  long long ans = 0;
  for (long long i = (1); i < (n - 1); ++i) {
    ans += query2(1, 0, n - 1, 0, i - 1, a[i]) *
           (n - 1 - i - query2(1, 0, n - 1, i + 1, n - 1, a[i]));
  }
  cout << ans;
}

### Prompt

Develop a solution in CPP 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;
vector<int> a(1000001);
vector<vector<int> > seg(4 * 1000001);
vector<int> merg(int node) {
  int l = 2 * node, r = 2 * node + 1;
  int sz1 = seg[l].size();
  int sz2 = seg[r].size();
  int k = 0, i = 0, j = 0;
  while (i < sz1 && j < sz2) {
    if (seg[l][i] < seg[r][j])
      seg[node].push_back(seg[l][i++]);
    else
      seg[node].push_back(seg[r][j++]);
  }
  if (i < sz1) {
    while (i < sz1) {
      seg[node].push_back(seg[l][i++]);
    }
  }
  if (j < sz2) {
    while (j < sz2) {
      seg[node].push_back(seg[r][j++]);
    }
  }
  return seg[node];
}
void build(int cur, int st, int end) {
  if (st == end) {
    vector<int> temp;
    temp.push_back(a[st]);
    seg[cur] = temp;
    return;
  }
  long long mid = (st + end) >> 1;
  build(2 * cur, st, mid);
  build(2 * cur + 1, mid + 1, end);
  merg(cur);
}
int query2(int cur, int st, int end, int l, int r, int k) {
  if (l <= st && r >= end) {
    int temp =
        lower_bound(seg[cur].begin(), seg[cur].end(), k) - seg[cur].begin();
    return seg[cur].size() - temp;
  }
  if (r < st || l > end) return 0;
  long long mid = (st + end) >> 1;
  long long ans1 = query2(2 * cur, st, mid, l, r, k);
  long long ans2 = query2(2 * cur + 1, mid + 1, end, l, r, k);
  return (ans1 + ans2);
}
int main() {
  iostream::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (long long i = (0); i < (n); ++i) cin >> a[i];
  build(1, 0, n - 1);
  long long ans = 0;
  for (long long i = (1); i < (n - 1); ++i) {
    ans += query2(1, 0, n - 1, 0, i - 1, a[i]) *
           (n - 1 - i - query2(1, 0, n - 1, i + 1, n - 1, a[i]));
  }
  cout << ans;
}

```