Play With Numbers Programming Challenge (finding mean value of subarray in java)

I'm trying to solve this Play With Numbers. I have passed the test cases but, I kept getting time limit exceeded. Can someone help me improve its performance in order to pass the time limit, please?

Problem:

You are given an array of n numbers and q queries. For each query you have to print the floor of the expected value(mean) of the subarray from L to R.

Input:

First line contains two integers N and Q denoting number of array elements and number of queries.

Next line contains N space seperated integers denoting array elements.

Next Q lines contain two integers L and R(indices of the array).

Output:

print a single integer denoting the answer.

Constraints:

1<= N ,Q,L,R <= 10^6

1<= Array elements <= 10^9

My code:

import java.io.*;
import java.util.*;

class TestClass 
{
    public static void main(String args[] ) throws IOException 
    {
       Scanner sc=new Scanner(System.in);
       int n= sc.nextInt();
       int q= sc.nextInt();
       int arr[]=new int[n];
       int sums[]=new int[n+1];
       sums[0]=0;
       for(int i=0;i<n;i++)
       {
           arr[i]=sc.nextInt();
           sums[i+1]=sums[i]+arr[i];

       }
       for(int i=0;i<q;i++)
       {
           
           int q1=sc.nextInt();
           int q2=sc.nextInt();
           int end=(q2-q1)+1;
           int mean= (sums[q2]-sums[q1-1])/end;
           System.out.println(mean);
       }
    }
}

According to my not-so-scientific tests, this doubles the speed:

    sc.nextLine(); // to finish row #1
    String line = sc.nextLine();
    String[] parts = line.split(" ");
    for (int i = 0; i < parts.length; i++) {
        arr[i] = Integer.parseInt(parts[i]);
        sums[i + 1] = sums[i] + arr[i];
    }

As you see, the little trick is to minimize the number of reads that the Scanner has to to (as there normally is a bit of 'lag' surrounding each 'read'). I read from a file, not from System.in, so maybe you will not see the same improvement in your setup.