hdu2601

/*
题意：给n 求满足i*j+i+j=n(0<i<=j)方案数

思路：xy+x+y=n
          (x+1)(y+1)=x*y+x+y+1=n+1;

          即求n+1的因子对数
参考：https://blog.csdn.net/Puppettt/article/details/83030925?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522172096421916800182136946%2522%252C%2522scm%2522%253A%252220140713.130102334..%2522%257D&request_id=172096421916800182136946&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2~all~sobaiduend~default-2-83030925-null-null.142^v100^pc_search_result_base8&utm_term=hdu2601&spm=1018.2226.3001.4187
*/

import java.util.Scanner;

public class hdu2601 {

    public static void main(String[] args) {
        // TODO 自动生成的方法存根
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            double num = sc.nextDouble();
            num++;
            int ans = 0;
            int cs;
            cs = (int) Math.sqrt(num);
            for (int j = 2; j <= cs; j++) {
                if (num%j==0) {
                    ans++;
                }                
            }
            System.out.println(ans);
        }
        sc.close();
    }

}