Problem:
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The Number of Inversions

For a given sequence $A = \{a_0, a_1, ... a_{n-1}\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:

bubbleSort(A)
cnt = 0 // the number of inversions
for i = 0 to A.length-1
for j = A.length-1 downto i+1
if A[j] < A[j-1]
swap(A[j], A[j-1])
cnt++

return cnt

For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.

Input

In the first line, an integer $n$, the number of elements in $A$, is given. In the second line, the elements $a_i$ ($i = 0, 1, .. n-1$) are given separated by space characters.

output

Print the number of inversions in a line.

Constraints

$ 1 \leq n \leq 200,000$

$ 0 \leq a_i \leq 10^9$

$a_i$ are all different

Sample Input 1

5
3 5 2 1 4

Sample Output 1

6

Sample Input 2

3
3 1 2

Sample Output 2

2

Solution in C++14 (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <utility>
#include <algorithm>

using namespace std;

long long inversion_count(vector<long long> &A) {
    auto N = (int) A.size();

    vector<pair<long long, int>> B(N);
    for (int i = 0; i < N; i++) B[i] = make_pair(A[i], i);
    sort(B.begin(), B.end());

    vector<int> idxs(N);
    for (int i = 0; i < N; i++) idxs[B[i].second] = i + 1;

    vector<int> BIT(N+1, 0);
    auto sum = [&](int i) {
        long long s = 0;
        while (i > 0) {
            s += BIT[i];
            i -= i & -i;
        }
        return s;
    };
    auto add_to = [&](int i, int x) {
        while (i <= N) {
            BIT[i] += x;
            i += i & -i;
        }
    };

    long long ret = 0;
    for (int j = 0; j < N; j++) {
        ret += j - sum(idxs[j]);
        add_to(idxs[j], 1);
    }

    return ret;
}

// Validation: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_5_D
signed main() {
    int N;
    cin >> N;
    vector<long long> A(N);
    for (int i = 0; i < N; i++) cin >> A[i];
    cout << inversion_count(A) << endl;
    return 0;
}