description: Polycarp must pay exactly n burles at the checkout. He has coins of two nominal values: 1 burle and 2 burles. Polycarp likes both kinds of coins equally. So he doesn't want to pay with more coins of one type than with the other.
  
  Thus, Polycarp wants to minimize the difference between the count of coins of 1 burle and 2 burles being used. Help him by determining two non-negative integer values c_1 and c_2 which are the number of coins of 1 burle and 2 burles, respectively, so that the total value of that number of coins is exactly n (i. e. c_1 + 2 ⋅ c_2 = n), and the absolute value of the difference between c_1 and c_2 is as little as possible (i. e. you must minimize |c_1-c_2|).
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line. This line contains one integer n (1 ≤ n ≤ 10^9) — the number of burles to be paid by Polycarp.
  
  Output
  
  For each test case, output a separate line containing two integers c_1 and c_2 (c_1, c_2 ≥ 0) separated by a space where c_1 is the number of coins of 1 burle and c_2 is the number of coins of 2 burles. If there are multiple optimal solutions, print any one.
  
  Example
  
  Input
  
  
  6
  1000
  30
  1
  32
  1000000000
  5
  
  
  Output
  
  
  334 333
  10 10
  1 0
  10 11
  333333334 333333333
  1 2
  
  Note
  
  The answer for the first test case is "334 333". The sum of the nominal values of all coins is 334 ⋅ 1 + 333 ⋅ 2 = 1000, whereas |334 - 333| = 1. One can't get the better value because if |c_1 - c_2| = 0, then c_1 = c_2 and c_1 ⋅ 1 + c_1 ⋅ 2 = 1000, but then the value of c_1 isn't an integer.
  
  The answer for the second test case is "10 10". The sum of the nominal values is 10 ⋅ 1 + 10 ⋅ 2 = 30 and |10 - 10| = 0, whereas there's no number having an absolute value less than 0.
solution: #include <bits/stdc++.h>
  int main() {
    int t, i;
    scanf("%d", &t);
    long n[t], n1[t], n2[t];
    for (i = 0; i < t; i++) {
      scanf("%ld", &n[i]);
      if ((n[i] % 3) == 1) {
        n1[i] = (n[i] / 3) + 1;
        n2[i] = n1[i] - 1;
      } else if ((n[i] % 3) == 2) {
        n1[i] = n[i] / 3;
        n2[i] = n1[i] + 1;
      } else if ((n[i] % 3) == 0) {
        n1[i] = n[i] / 3;
        n2[i] = n1[i];
      }
    }
    for (i = 0; i < t; i++) printf("%ld %ld\n", n1[i], n2[i]);
  }
  