Think of a whole number between one and a thousand. Let\'s go with 10 for now.

How many odd numbers are there between 1 and 10, (those numbers that are not divisible by 2)? There are five: 1, 3, 5, 7 and 9.

Using only these odd numbers, I want to find out the number of ways I can sum up these numbers such that I get a sum of 10. BUT there need to be five terms in my series of sums. I am allowed to re-use an odd number from my set more than once.

There aren\'t any. However, if the required number of terms was reduced to four, there are three sums that can form in this manner, namely:

1+1+3+5=10

1+3+3+3=10

1+1+1+7=10

What if I picked another number between 1 and 1000 as my sum? And what if the number of odd-numbered terms to form that sum was N?

Write a program that tells how many sums there are.

Input

Two integers on one line separated by a space.

N is the number of terms required for the sum, based on odd numbers in the range {1 to S}, if S is odd, or the range {1 to S - 1}, if S is even (1 =>=>

S is the sum to be formed from the specified odd number set (1 =>=>

Output

A single integer, denoting the number of sums that can be produced.

Sample Input

4 10

Sample Output

3

Code given:

public class OddNumber {

public static void main(String[] args) {

}

}

Sol155:

import java.util.Scanner;

public class OddNumber { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); int S = sc.nextInt(); int[] odd = new int[S/2]; int count = 0; for (int i = 0; i = 0; j--) { indices[j] = temp % odd.length; temp /= odd.length; } int sum = 0; for (int j = 0; j

The program takes two integer inputs from the user - N and S - which represent the number of odd terms required to form the sum and the sum itself, respectively. The program then generates an array of odd numbers from 1 to S, and uses nested loops to generate all possible combinations of N terms from this array. It then checks each combination to see if its sum is equal to S, and increments a count variable if it is. Finally, the program outputs the count of valid combinations.

Note that for large values of N and S, this program may take a very long time to run. However, for the given constraints of 1 =>=>=>=>