#include <bits/stdc++.h>
using namespace std;
int a[1000004];
map<int, int> op;
int b[1000005];
int seg1[4 * 1000004];
int seg2[4 * 1000004];
void build(int l, int r, int p) {
  if (l == r) {
    seg1[p] = op[l];
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, 2 * p);
  build(mid + 1, r, 2 * p + 1);
  seg1[p] = seg1[2 * p] + seg1[2 * p + 1];
}
void upd(int l, int r, int p, int idx) {
  if (l == r) {
    seg1[p] = 0;
    seg2[p] = 1;
    return;
  } else {
    int mid = (l + r) / 2;
    if (l <= idx && idx <= mid)
      upd(l, mid, 2 * p, idx);
    else
      upd(mid + 1, r, 2 * p + 1, idx);
    seg1[p] = seg1[2 * p] + seg1[2 * p + 1];
    seg2[p] = seg2[2 * p] + seg2[2 * p + 1];
  }
}
int quere1(int p, int start, int endd, int l, int r) {
  if (r < start || endd < l) return 0;
  if (l <= start && endd <= r) {
    return seg1[p];
  }
  int mid = (start + endd) / 2;
  int p1 = quere1(2 * p, start, mid, l, r);
  int p2 = quere1(2 * p + 1, mid + 1, endd, l, r);
  return (p1 + p2);
}
int quere2(int p, int start, int endd, int l, int r) {
  if (r < start || endd < l) return 0;
  if (l <= start && endd <= r) {
    return seg2[p];
  }
  int mid = (start + endd) / 2;
  int p1 = quere2(2 * p, start, mid, l, r);
  int p2 = quere2(2 * p + 1, mid + 1, endd, l, r);
  return (p1 + p2);
}
int main() {
  int n;
  cin >> n;
  int i;
  for (i = 0; i < n; i++) cin >> a[i], b[i] = a[i];
  sort(b, b + n);
  int cnt = 1;
  for (i = 0; i < n; i++) {
    op[b[i]] = cnt;
    cnt++;
  }
  for (i = 0; i < n; i++) a[i] = op[a[i]];
  op.clear();
  for (i = 0; i < n; i++) op[a[i]]++;
  build(0, n, 1);
  long long res = 0;
  for (i = 0; i < n; i++) {
    int x = a[i];
    long long c1 = quere1(1, 0, n, 1, x - 1);
    long long c2 = quere2(1, 0, n, x + 1, n);
    upd(0, n, 1, x);
    res += (c1 * c2);
  }
  cout << res;
  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;
int a[1000004];
map<int, int> op;
int b[1000005];
int seg1[4 * 1000004];
int seg2[4 * 1000004];
void build(int l, int r, int p) {
  if (l == r) {
    seg1[p] = op[l];
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, 2 * p);
  build(mid + 1, r, 2 * p + 1);
  seg1[p] = seg1[2 * p] + seg1[2 * p + 1];
}
void upd(int l, int r, int p, int idx) {
  if (l == r) {
    seg1[p] = 0;
    seg2[p] = 1;
    return;
  } else {
    int mid = (l + r) / 2;
    if (l <= idx && idx <= mid)
      upd(l, mid, 2 * p, idx);
    else
      upd(mid + 1, r, 2 * p + 1, idx);
    seg1[p] = seg1[2 * p] + seg1[2 * p + 1];
    seg2[p] = seg2[2 * p] + seg2[2 * p + 1];
  }
}
int quere1(int p, int start, int endd, int l, int r) {
  if (r < start || endd < l) return 0;
  if (l <= start && endd <= r) {
    return seg1[p];
  }
  int mid = (start + endd) / 2;
  int p1 = quere1(2 * p, start, mid, l, r);
  int p2 = quere1(2 * p + 1, mid + 1, endd, l, r);
  return (p1 + p2);
}
int quere2(int p, int start, int endd, int l, int r) {
  if (r < start || endd < l) return 0;
  if (l <= start && endd <= r) {
    return seg2[p];
  }
  int mid = (start + endd) / 2;
  int p1 = quere2(2 * p, start, mid, l, r);
  int p2 = quere2(2 * p + 1, mid + 1, endd, l, r);
  return (p1 + p2);
}
int main() {
  int n;
  cin >> n;
  int i;
  for (i = 0; i < n; i++) cin >> a[i], b[i] = a[i];
  sort(b, b + n);
  int cnt = 1;
  for (i = 0; i < n; i++) {
    op[b[i]] = cnt;
    cnt++;
  }
  for (i = 0; i < n; i++) a[i] = op[a[i]];
  op.clear();
  for (i = 0; i < n; i++) op[a[i]]++;
  build(0, n, 1);
  long long res = 0;
  for (i = 0; i < n; i++) {
    int x = a[i];
    long long c1 = quere1(1, 0, n, 1, x - 1);
    long long c2 = quere2(1, 0, n, x + 1, n);
    upd(0, n, 1, x);
    res += (c1 * c2);
  }
  cout << res;
  return 0;
}

```