This problem gives you a flavor of the concept of a special judge. That means the judge is smart enough to verify your code even though it may print different results. In this problem, you are asked to find the opposite task of the previous problem.

To be specific, I have two computers where I stored my problems. Now I know the total number of problems is n. There are no duplicate problems, and there can be at most 10 problems in each computer. You have to find the number of problems on each of the computers.

Since there can be multiple solutions. Any valid solution will do.

Input

Input starts with an integer T (≤ 25), denoting the number of test cases.

Each case starts with a line containing an integer n (0 ≤ n ≤ 20) denoting the total number of problems.

Output

For each case, print the number of problems stored in each computer in a single line. A single space should separate the non-negative integers.

Sample

Sample Input	Sample Output
Click to copy 3 10 7 7	Click to copy 0 10 0 7 1 6

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LOJ 1001 - Opposite Task

In this problem, you are given T test cases. Each test case contains an integer n, the value of which is 20 at max. Your task is to print out two non-negative integers a and b, such that n = a + b. Any values of a and b that satisfies the equation is accepted, however, neither of a nor b can be greater than 10.

It is one of the easiest problems on LightOJ volumes. However, some people might find it difficult to get an Accepted verdict. The reason could be one of the following:

Printing a number greater than 10
Failing to provide proper output format
Missing new line on each line

If you are still stuck with this problem, check the codes below:

C

#include <stdio.h>
int main() {
  int T, n;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) {
    scanf("%d", &n);
    if (n > 10) {
      printf("10 %d\n", n - 10);
    } else {
      printf("0 %d\n", n);
    }
  }
  return 0;
}

Java

package com.loj.volume;

import java.util.Scanner;

class OppositeTask {
  public static void main(String[] args) throws Exception {
    Scanner scanner = new Scanner(System.in);
    int T = scanner.nextInt(), n;
    for (int i = 1; i <= T; i++) {
      n = scanner.nextInt();
      if (n > 10) {
        System.out.printf("10 %d\n", n - 10);
      } else {
        System.out.printf("0 %d\n", n);
      }
    }
  }
}

Tutorial source