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 JAVA (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;
 class FS {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        return hasNextByte() ? buffer[ptr++] : -1;
    }

    private boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
    }

    public boolean hasNext() {
        skipUnprintable();
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public int nextInt() {
        return (int) nextLong();
    }

    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
}
public class Main {
    static long k = 0;

    public static void main(String[] args) {
        FS fs = new FS();
        int n = fs.nextInt();
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = fs.nextInt();
        }
        mergeSort(a, 0, a.length);
        System.out.println(k);
    }

    private static void mergeSort(int[] a, int left, int right) {
        if (left+1 < right) {
            int mid = (left + right) / 2;
            mergeSort(a, left, mid);
            mergeSort(a, mid, right);
            merge(a, left, mid, right);
        }
    }

    private static void merge(int[] a, int left, int mid, int right) {
        int n1 = mid - left;
        int n2 = right - mid;
        int l[] = new int[n1+1];
        int r[] = new int[n2+1];
        for (int i = 0; i < n1; i++) {
            l[i]=a[left+i];
        }
        for (int i = 0; i < n2; i++) {
            r[i]=a[mid+i];
        }
        l[n1]=Integer.MAX_VALUE;
        r[n2]=Integer.MAX_VALUE;
        int m=0,n=0;
        for (int i = left; i < right; i++) {
            if (l[m]<=r[n]){
                a[i]=l[m];
                m++;
            }else{
                a[i]=r[n];
                n++;
                k+=(n1-m);
            }
        }
    }
}