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;

public class Main {
	private static FastScanner sc = new FastScanner();
	static long cnt = 0;

	public static void main(String[] args) {
		int n = sc.nextInt();
		
		int[] A = new int[n];
		for(int i=0; i<n; i++) {
			A[i] = sc.nextInt();
		}
		
		mergeSort(A, 0, n);
		
		System.out.println(cnt);
	}
	
	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 i = 0;
		int j = 0;
		for(int k=left; k<right; k++) {
			if(L[i] <= R[j]) {
				A[k] = L[i];
				i++;
			} else {
				A[k] = R[j];
				j++;
				cnt += n1 - i;
			}
		}
	}
	
	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);
		}
	}
	
	static class FastScanner {
	    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() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
	    private static 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 long nextLong() {
	        return Long.parseLong(next());
	    }
	    public int nextInt(){
	    	return Integer.parseInt(next());
	    }
	}
}