description: Given a positive integer n. Find three distinct positive integers a, b, c such that a + b + c = n and \operatorname{gcd}(a, b) = c, where \operatorname{gcd}(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
  
  The first and only line of each test case contains a single integer n (10 ≤ n ≤ 10^9).
  
  Output
  
  For each test case, output three distinct positive integers a, b, c satisfying the requirements. If there are multiple solutions, you can print any. We can show that an answer always exists.
  
  Example
  
  Input
  
  
  6
  18
  63
  73
  91
  438
  122690412
  
  
  Output
  
  
  6 9 3
  21 39 3
  29 43 1
  49 35 7
  146 219 73
  28622 122661788 2
  
  Note
  
  In the first test case, 6 + 9 + 3 = 18 and \operatorname{gcd}(6, 9) = 3.
  
  In the second test case, 21 + 39 + 3 = 63 and \operatorname{gcd}(21, 39) = 3.
  
  In the third test case, 29 + 43 + 1 = 73 and \operatorname{gcd}(29, 43) = 1.
solution: #include <bits/stdc++.h>
  using namespace std;
  constexpr long long bits(long long x) {
    return x == 0 ? 0 : 31 - __builtin_clz(x);
  }
  long long INF = 2000000007;
  long long INFLL = (long long)INF * (long long)INF;
  double EPS = 10e-7;
  long double pi = 2 * acos(0.0);
  long long mod = 998244353;
  const long long array_size = 2e5 + 7;
  long long Power(long long aa, long long bb) {
    long long $a = aa;
    long long $b = bb;
    long long $result = 1ll;
    $a %= 1000000007;
    while ($b) {
      if ($b % 2ll == 1ll) $result *= $a;
      $a *= $a;
      $a %= 1000000007;
      $result %= 1000000007;
      $b /= 2ll;
    }
    return $result % 1000000007;
  }
  void solve(long long tc) {
    long long n;
    cin >> n;
    if (n == 10) {
      cout << "7 2 1" << endl;
      return;
    }
    if (n % 2 == 0) {
      cout << (n - 1) / 2 << " " << (n + 1) / 2 << " " << 1 << endl;
      return;
    }
    n -= 1;
    n /= 2;
    if (n % 2 == 0) {
      cout << n - 1 << " " << n + 1 << " ";
      cout << 1 << endl;
      return;
    }
    cout << n - 2 << " " << n + 2 << " ";
    cout << 1 << endl;
  }
  int32_t main() {
    clock_t start, end;
    start = clock();
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long tc = 1;
    cin >> tc;
    for (long long i = 1; i <= tc; i++) {
      solve(i);
    }
    end = clock();
    double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
    cerr << "\nTime taken by program is : " << fixed << time_taken
         << setprecision(5);
    cerr << " sec " << endl;
    return 0;
  }
  