#include <bits/stdc++.h>
using namespace std;
struct node {
  int val;
  int pos;
};
int n;
unsigned long long f[1000001], bit[1000001];
node a[1000001];
bool cmp(node x, node y) { return x.val > y.val; }
void build_bit() {
  for (int i = 1; i <= n; i++) bit[i] = i & -i;
}
void update(int x, unsigned long long val) {
  while (x <= n) {
    bit[x] += 1ull * val;
    x += x & -x;
  }
}
unsigned long long sum(int x) {
  unsigned long long ans = 0;
  while (x > 0) {
    ans += 1ull * bit[x];
    x -= x & -x;
  }
  return ans;
}
void init() {
  int j = 1, k = 1, ans = 0;
  build_bit();
  for (int i = 1; i <= n; i++) {
    while (j <= i && a[j].val >= a[i].val) {
      update(a[j].pos, -1);
      j++;
    }
    f[a[i].pos] = sum(n) - sum(a[i].pos);
  }
}
void doit() {
  unsigned long long s[n + 1], ans = 0;
  s[0] = 0;
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + 1ull * f[i];
  for (int i = 1; i <= n; i++) bit[i] = s[i] - s[i - (i & -i)];
  int j = 1;
  for (int i = 1; i <= n; i++) {
    while (j <= i && a[j].val >= a[i].val) {
      update(a[j].pos, -f[a[j].pos]);
      j++;
    }
    ans = ans + 1ull * sum(n) - 1ull * sum(a[i].pos);
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0), cout.tie(0);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].val);
    a[i].pos = i;
  }
  sort(a + 1, a + 1 + n, &cmp);
  init();
  doit();
  return 0;
}

### Prompt

Please formulate 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;
struct node {
  int val;
  int pos;
};
int n;
unsigned long long f[1000001], bit[1000001];
node a[1000001];
bool cmp(node x, node y) { return x.val > y.val; }
void build_bit() {
  for (int i = 1; i <= n; i++) bit[i] = i & -i;
}
void update(int x, unsigned long long val) {
  while (x <= n) {
    bit[x] += 1ull * val;
    x += x & -x;
  }
}
unsigned long long sum(int x) {
  unsigned long long ans = 0;
  while (x > 0) {
    ans += 1ull * bit[x];
    x -= x & -x;
  }
  return ans;
}
void init() {
  int j = 1, k = 1, ans = 0;
  build_bit();
  for (int i = 1; i <= n; i++) {
    while (j <= i && a[j].val >= a[i].val) {
      update(a[j].pos, -1);
      j++;
    }
    f[a[i].pos] = sum(n) - sum(a[i].pos);
  }
}
void doit() {
  unsigned long long s[n + 1], ans = 0;
  s[0] = 0;
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + 1ull * f[i];
  for (int i = 1; i <= n; i++) bit[i] = s[i] - s[i - (i & -i)];
  int j = 1;
  for (int i = 1; i <= n; i++) {
    while (j <= i && a[j].val >= a[i].val) {
      update(a[j].pos, -f[a[j].pos]);
      j++;
    }
    ans = ans + 1ull * sum(n) - 1ull * sum(a[i].pos);
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0), cout.tie(0);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].val);
    a[i].pos = i;
  }
  sort(a + 1, a + 1 + n, &cmp);
  init();
  doit();
  return 0;
}

```