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;
  void input(vector<long long int> &arr) {
    for (int i = 0; i < arr.size(); i++) cin >> arr[i];
  }
  void output(vector<long long int> &arr) {
    for (int i = 0; i < arr.size(); i++) cout << arr[i] << " ";
    cout << "\n";
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long int power(long long int a, long long int b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    long long int smallans = power(a, b / 2);
    long long int myans = (smallans * smallans) % 1000000007;
    if ((b & 1)) myans = (myans * a) % 1000000007;
    return myans;
  }
  long long int multiply(long long int a, long long int b) {
    long long int ans = ((a % 1000000007) * (b % 1000000007)) % 1000000007;
    return ans;
  }
  long long int divide(long long int a, long long int b) {
    return multiply(a, power(b, 1000000007 - 2));
  }
  void manageLazy(vector<long long int> &tree, vector<long long int> &lazy,
                  long long int idx) {
    tree[2 * idx + 1] += lazy[idx];
    lazy[2 * idx + 1] += lazy[idx];
    tree[2 * idx + 2] += lazy[idx];
    lazy[2 * idx + 2] += lazy[idx];
    lazy[idx] = 0;
  }
  void build(vector<long long int> &tree, vector<long long int> &arr,
             long long int left, long long int right, long long int idx) {
    if (left == right) {
      tree[idx] = arr[left];
      return;
    }
    long long int mid = (left + right) / 2;
    build(tree, arr, left, mid, 2 * idx + 1);
    build(tree, arr, mid + 1, right, 2 * idx + 2);
    tree[idx] = min(tree[2 * idx + 1], tree[2 * idx + 2]);
  }
  void update(vector<long long int> &tree, vector<long long int> &lazy,
              long long int tl, long long int tr, long long int l,
              long long int r, long long int idx, long long int val) {
    if (l > r) return;
    if (l == tl && tr == r) {
      tree[idx] += val;
      lazy[idx] += val;
      return;
    }
    manageLazy(tree, lazy, idx);
    long long int mid = (tl + tr) / 2;
    update(tree, lazy, tl, mid, l, min(r, mid), 2 * idx + 1, val);
    update(tree, lazy, mid + 1, tr, max(l, mid + 1), r, 2 * idx + 2, val);
    tree[idx] = min(tree[2 * idx + 1], tree[2 * idx + 2]);
  }
  long long int query(vector<long long int> &tree, vector<long long int> &lazy,
                      long long int tl, long long int tr, long long int l,
                      long long int r, long long int idx) {
    if (l > r) return 1e18;
    if (l <= tl && tr <= r) return tree[idx];
    manageLazy(tree, lazy, idx);
    long long int mid = (tl + tr) / 2;
    long long int a = query(tree, lazy, tl, mid, l, min(r, mid), 2 * idx + 1);
    long long int b =
        query(tree, lazy, mid + 1, tr, max(l, mid + 1), r, 2 * idx + 2);
    return min(a, b);
  }
  void update(vector<long long int> &tree, long long int index,
              long long int val) {
    index++;
    while (index < tree.size()) {
      tree[index] = (tree[index] + val) % 1000000007;
      index += index & (-index);
    }
  }
  long long int calculate(vector<long long int> &tree, long long int index) {
    long long int sum = 0;
    index++;
    while (index > 0) {
      sum = (sum + tree[index]) % 1000000007;
      index -= index & (-index);
    }
    return sum;
  }
  long long int findpar(long long int p, vector<long long int> &parent) {
    if (parent[p] == p) return p;
    parent[p] = findpar(parent[p], parent);
    return parent[p];
  }
  void merge(long long int a, long long int b, vector<long long int> &parent,
             vector<long long int> &size_) {
    a = findpar(a, parent);
    b = findpar(b, parent);
    if (a != b) {
      if (size_[a] < size_[b]) swap(a, b);
      parent[b] = a;
      size_[a] += size_[b];
    }
  }
  void sieve(vector<bool> &primes, vector<long long int> &count) {
    for (int i = 2; i < primes.size(); i++) {
      if (primes[i]) {
        count.push_back(i);
        for (int j = i * i; j < primes.size(); j += i) primes[j] = false;
      }
    }
  }
  void solve() {
    long long int n;
    cin >> n;
    long long int a, b, c;
    c = 1;
    long long int t = n - 1;
    if (t % 2) {
      cout << 2 << " " << t - 2 << " " << 1 << "\n";
      return;
    }
    if ((t / 2) % 2 == 0)
      cout << t / 2 + 1 << " " << t / 2 - 1 << " " << 1 << "\n";
    else
      cout << t / 2 + 2 << " " << t / 2 - 2 << " " << 1 << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  