#include <bits/stdc++.h>
using namespace std;
const long long mxn = 1e6 + 2;
const long long inf = 1e9 + 7;
const long long mod = 1e9 + 7;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
map<long long, long long> mp;
long long n, a[mxn], b[mxn], c[mxn], seg[4 * mxn], sum[mxn], sum2[mxn], num = 0;
void build(long long v, long long tl, long long tr) {
  if (tl == tr) {
    seg[v] = 0;
  } else {
    long long tm = (tl + tr) / 2;
    build((v << 1), tl, tm);
    build(((v << 1) | 1), tm + 1, tr);
    seg[v] = seg[((v << 1) | 1)] + seg[(v << 1)];
  }
}
void add(long long v, long long tl, long long tr, long long pos) {
  if (tl == tr && tl == pos) {
    seg[v] += 1;
    c[tl] += 1;
  } else {
    long long tm = (tl + tr) / 2;
    if (pos <= tm)
      add((v << 1), tl, tm, pos);
    else
      add(((v << 1) | 1), tm + 1, tr, pos);
    seg[v]++;
  }
}
long long ans(long long v, long long tl, long long tr, long long l,
              long long r) {
  if (l > r) return 0;
  if (tl == l && tr == r) {
    return seg[v];
    c[tl]++;
  } else {
    long long tm = (tl + tr) / 2;
    return ans((v << 1), tl, tm, l, min(tm, r)) +
           ans(((v << 1) | 1), tm + 1, tr, max(l, tm + 1), r);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (b[i] != b[i - 1] || i == 1) {
      mp[b[i]] = i;
    }
  }
  for (int i = 1; i <= n; i++) {
    sum[i] += ans(1, 1, n, mp[a[i]] + 1, n);
    add(1, 1, n, mp[a[i]]);
  }
  reverse(a + 1, a + 1 + n);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    sum2[n - i + 1] += ans(1, 1, n, 1, mp[a[i]] - 1);
    add(1, 1, n, mp[a[i]]);
  }
  for (int i = 1; i <= n; i++) num += sum[i] * sum2[i];
  cout << num << endl;
  return 0;
}

### Prompt

Generate 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 long long mxn = 1e6 + 2;
const long long inf = 1e9 + 7;
const long long mod = 1e9 + 7;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
map<long long, long long> mp;
long long n, a[mxn], b[mxn], c[mxn], seg[4 * mxn], sum[mxn], sum2[mxn], num = 0;
void build(long long v, long long tl, long long tr) {
  if (tl == tr) {
    seg[v] = 0;
  } else {
    long long tm = (tl + tr) / 2;
    build((v << 1), tl, tm);
    build(((v << 1) | 1), tm + 1, tr);
    seg[v] = seg[((v << 1) | 1)] + seg[(v << 1)];
  }
}
void add(long long v, long long tl, long long tr, long long pos) {
  if (tl == tr && tl == pos) {
    seg[v] += 1;
    c[tl] += 1;
  } else {
    long long tm = (tl + tr) / 2;
    if (pos <= tm)
      add((v << 1), tl, tm, pos);
    else
      add(((v << 1) | 1), tm + 1, tr, pos);
    seg[v]++;
  }
}
long long ans(long long v, long long tl, long long tr, long long l,
              long long r) {
  if (l > r) return 0;
  if (tl == l && tr == r) {
    return seg[v];
    c[tl]++;
  } else {
    long long tm = (tl + tr) / 2;
    return ans((v << 1), tl, tm, l, min(tm, r)) +
           ans(((v << 1) | 1), tm + 1, tr, max(l, tm + 1), r);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (b[i] != b[i - 1] || i == 1) {
      mp[b[i]] = i;
    }
  }
  for (int i = 1; i <= n; i++) {
    sum[i] += ans(1, 1, n, mp[a[i]] + 1, n);
    add(1, 1, n, mp[a[i]]);
  }
  reverse(a + 1, a + 1 + n);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    sum2[n - i + 1] += ans(1, 1, n, 1, mp[a[i]] - 1);
    add(1, 1, n, mp[a[i]]);
  }
  for (int i = 1; i <= n; i++) num += sum[i] * sum2[i];
  cout << num << endl;
  return 0;
}

```