#include <bits/stdc++.h>
using namespace std;
vector<long long> sums;
struct segtree {
  long long size = 1;
  void init(long long n) {
    while (size < n) {
      size *= 2;
    }
    sums.assign(2 * size, 0);
  }
  void set(long long i, long long v, long long x, long long lx, long long rx) {
    if (rx - lx == 1) {
      sums[x] = v;
      return;
    }
    long long m = (lx + rx) / 2;
    if (i < m) {
      set(i, v, 2 * x + 1, lx, m);
    } else {
      set(i, v, 2 * x + 2, m, rx);
    }
    sums[x] = sums[2 * x + 1] + sums[2 * x + 2];
  }
  void set(long long i, long long v) { set(i, v, 0, 0, size); }
  long long sum(long long l, long long r, long long x, long long lx,
                long long rx) {
    if (l >= rx || r <= lx) return 0;
    if (lx >= l && rx <= r) return sums[x];
    long long m = (lx + rx) / 2;
    long long s1 = sum(l, r, 2 * x + 1, lx, m);
    long long s2 = sum(l, r, 2 * x + 2, m, rx);
    return s1 + s2;
  }
  long long sum(long long l, long long r) { return sum(l, r, 0, 0, size); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  segtree st;
  st.init(n);
  long long a[n], b[n], p1[n], p2[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
    st.set(i, 0);
  }
  sort(b, b + n);
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) m[b[i]] = i;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    st.set(m[a[i]], 1);
    p1[i] = st.sum(m[a[i]] + 1, n);
  }
  for (long long i = 0; i < n; i++) {
    st.set(m[a[i]], 0);
  }
  for (long long i = n - 1; i >= 0; i--) {
    st.set(m[a[i]], 1);
    p2[i] = st.sum(0, m[a[i]]);
  }
  for (long long i = 0; i < n; i++) ans += (p1[i] * p2[i]);
  cout << ans << "\n";
}

### 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;
vector<long long> sums;
struct segtree {
  long long size = 1;
  void init(long long n) {
    while (size < n) {
      size *= 2;
    }
    sums.assign(2 * size, 0);
  }
  void set(long long i, long long v, long long x, long long lx, long long rx) {
    if (rx - lx == 1) {
      sums[x] = v;
      return;
    }
    long long m = (lx + rx) / 2;
    if (i < m) {
      set(i, v, 2 * x + 1, lx, m);
    } else {
      set(i, v, 2 * x + 2, m, rx);
    }
    sums[x] = sums[2 * x + 1] + sums[2 * x + 2];
  }
  void set(long long i, long long v) { set(i, v, 0, 0, size); }
  long long sum(long long l, long long r, long long x, long long lx,
                long long rx) {
    if (l >= rx || r <= lx) return 0;
    if (lx >= l && rx <= r) return sums[x];
    long long m = (lx + rx) / 2;
    long long s1 = sum(l, r, 2 * x + 1, lx, m);
    long long s2 = sum(l, r, 2 * x + 2, m, rx);
    return s1 + s2;
  }
  long long sum(long long l, long long r) { return sum(l, r, 0, 0, size); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  segtree st;
  st.init(n);
  long long a[n], b[n], p1[n], p2[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
    st.set(i, 0);
  }
  sort(b, b + n);
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) m[b[i]] = i;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    st.set(m[a[i]], 1);
    p1[i] = st.sum(m[a[i]] + 1, n);
  }
  for (long long i = 0; i < n; i++) {
    st.set(m[a[i]], 0);
  }
  for (long long i = n - 1; i >= 0; i--) {
    st.set(m[a[i]], 1);
    p2[i] = st.sum(0, m[a[i]]);
  }
  for (long long i = 0; i < n; i++) ans += (p1[i] * p2[i]);
  cout << ans << "\n";
}

```