#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
vector<long long> tree[4 * maxn], rr;
long long arr[maxn];
int bs(vector<long long> vec, long long key) {
  int lo = 0, hi = vec.size() - 1;
  while (lo <= hi) {
    long long mid = (lo + hi) >> 1;
    if (vec[mid] == key) {
      return mid + 1;
    } else if (vec[mid] > key)
      hi = mid - 1;
    else
      lo = mid + 1;
  }
}
void build(int node, int b, int e) {
  if (b == e) {
    tree[node].push_back(arr[b]);
    return;
  }
  int mid = (b + e) >> 1;
  build(2 * node, b, mid);
  build(2 * node + 1, mid + 1, e);
  merge(tree[2 * node].begin(), tree[2 * node].end(),
        tree[2 * node + 1].begin(), tree[2 * node + 1].end(),
        back_inserter(tree[node]));
}
int query(int type, int node, int b, int e, int i, int j, long long num) {
  if (i > e or j < b) return 0;
  if (b >= i and j >= e) {
    if (type == 0) {
      int v = upper_bound(tree[node].begin(), tree[node].end(), num) -
              lower_bound(tree[node].begin(), tree[node].end(), 0);
      return (int)tree[node].size() - v;
    } else {
      int v = lower_bound(tree[node].begin(), tree[node].end(), num) -
              lower_bound(tree[node].begin(), tree[node].end(), 0);
      return v;
    }
  }
  int mid = (b + e) >> 1;
  int p1 = query(type, 2 * node, b, mid, i, j, num);
  int p2 = query(type, 2 * node + 1, mid + 1, e, i, j, num);
  return p1 + p2;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  build(1, 1, n);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long ans1 = query(0, 1, 1, n, 1, i, arr[i]);
    long long ans2 = query(1, 1, 1, n, i, n, arr[i]);
    ans += ans1 * ans2;
  }
  cout << ans << endl;
}

### Prompt

Your challenge is to write a CPP solution to 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 maxn = 1e6 + 5;
vector<long long> tree[4 * maxn], rr;
long long arr[maxn];
int bs(vector<long long> vec, long long key) {
  int lo = 0, hi = vec.size() - 1;
  while (lo <= hi) {
    long long mid = (lo + hi) >> 1;
    if (vec[mid] == key) {
      return mid + 1;
    } else if (vec[mid] > key)
      hi = mid - 1;
    else
      lo = mid + 1;
  }
}
void build(int node, int b, int e) {
  if (b == e) {
    tree[node].push_back(arr[b]);
    return;
  }
  int mid = (b + e) >> 1;
  build(2 * node, b, mid);
  build(2 * node + 1, mid + 1, e);
  merge(tree[2 * node].begin(), tree[2 * node].end(),
        tree[2 * node + 1].begin(), tree[2 * node + 1].end(),
        back_inserter(tree[node]));
}
int query(int type, int node, int b, int e, int i, int j, long long num) {
  if (i > e or j < b) return 0;
  if (b >= i and j >= e) {
    if (type == 0) {
      int v = upper_bound(tree[node].begin(), tree[node].end(), num) -
              lower_bound(tree[node].begin(), tree[node].end(), 0);
      return (int)tree[node].size() - v;
    } else {
      int v = lower_bound(tree[node].begin(), tree[node].end(), num) -
              lower_bound(tree[node].begin(), tree[node].end(), 0);
      return v;
    }
  }
  int mid = (b + e) >> 1;
  int p1 = query(type, 2 * node, b, mid, i, j, num);
  int p2 = query(type, 2 * node + 1, mid + 1, e, i, j, num);
  return p1 + p2;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  build(1, 1, n);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long ans1 = query(0, 1, 1, n, 1, i, arr[i]);
    long long ans2 = query(1, 1, 1, n, i, n, arr[i]);
    ans += ans1 * ans2;
  }
  cout << ans << endl;
}

```