#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)1e11;
const int maxn = (int)1e6;
const int maxk = 10;
struct segment_tree {
  vector<vector<long long>> st;
  int ts;
  segment_tree() : ts(0) {}
  segment_tree(int n, int k) : ts(1) {
    while (ts < n) ts <<= 1;
    st.resize(2 * ts, vector<long long>(k));
  }
  long long get_f(int l, int r, int L, int R, int j = 0, int v = 1) {
    if (l <= L && R <= r) return st[v][j];
    if (l > R || r < L) return 0;
    return get_f(l, r, L, (L + R) / 2, j, 2 * v) +
           get_f(l, r, (L + R) / 2 + 1, R, j, 2 * v + 1);
  }
  void change_element(int i, int j = 0, long long val = 1) {
    for (i += ts - 1; i >= 1; i /= 2) st[i][j] += val;
  }
};
int power[maxn];
long long inverse[maxn][maxk];
pair<int, int> c[maxn];
int n, k;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  k = 3;
  for (int i = 0; i < n; i++) {
    cin >> c[i].first;
    c[i].second = i;
  }
  sort(c, c + n);
  for (int i = 0; i < n; i++) power[c[i].second] = i + 1;
  segment_tree st(n, k - 1);
  for (int j = 0; j < k - 1; j++) {
    for (int i = n - 1; i >= 0; i--) {
      if (j) st.change_element(power[i], j, inverse[i][j - 1]);
      inverse[i][j] = st.get_f(1, power[i] - 1, 1, st.ts, j);
      if (!j) st.change_element(power[i]);
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += inverse[i][k - 2];
  }
  cout << ans;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 INF = (long long)1e11;
const int maxn = (int)1e6;
const int maxk = 10;
struct segment_tree {
  vector<vector<long long>> st;
  int ts;
  segment_tree() : ts(0) {}
  segment_tree(int n, int k) : ts(1) {
    while (ts < n) ts <<= 1;
    st.resize(2 * ts, vector<long long>(k));
  }
  long long get_f(int l, int r, int L, int R, int j = 0, int v = 1) {
    if (l <= L && R <= r) return st[v][j];
    if (l > R || r < L) return 0;
    return get_f(l, r, L, (L + R) / 2, j, 2 * v) +
           get_f(l, r, (L + R) / 2 + 1, R, j, 2 * v + 1);
  }
  void change_element(int i, int j = 0, long long val = 1) {
    for (i += ts - 1; i >= 1; i /= 2) st[i][j] += val;
  }
};
int power[maxn];
long long inverse[maxn][maxk];
pair<int, int> c[maxn];
int n, k;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  k = 3;
  for (int i = 0; i < n; i++) {
    cin >> c[i].first;
    c[i].second = i;
  }
  sort(c, c + n);
  for (int i = 0; i < n; i++) power[c[i].second] = i + 1;
  segment_tree st(n, k - 1);
  for (int j = 0; j < k - 1; j++) {
    for (int i = n - 1; i >= 0; i--) {
      if (j) st.change_element(power[i], j, inverse[i][j - 1]);
      inverse[i][j] = st.get_f(1, power[i] - 1, 1, st.ts, j);
      if (!j) st.change_element(power[i]);
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += inverse[i][k - 2];
  }
  cout << ans;
  return 0;
}

```