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.
1
669278611
 10
10
52
91
23
44
99
28
54
99
51

334639303 334639307 1
 4 5 1
25 26 1
43 47 1
9 13 1
21 22 1
47 51 1
13 14 1
26 27 1
47 51 1
23 27 1

#include<bits/stdc++.h>
using namespace std;

#define ff              first
#define ss              second
#define all(v)          v.begin(), v.end()
#define int             long long
#define pii             pair<int,int>
#define vi              vector<int>
#define mod             1000000007
#define inf             1e18

#define deb(...) cerr << "[" << #__VA_ARGS__ << "]: "; cerr << to_string(__VA_ARGS__) << endl
template <typename T, size_t N> int SIZE(const T (&t)[N]) { return N; } template<typename T> int SIZE(const T &t) { return t.size(); } string to_string(const string s, int x1 = 0, int x2 = 1e9) { return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"'; } string to_string(const char* s) { return to_string((string) s); } string to_string(const bool b) { return (b ? "true" : "false"); } string to_string(const char c) { return string({c}); } template<size_t N> string to_string(const bitset<N> &b, int x1 = 0, int x2 = 1e9) { string t = ""; for (int __iii__ = min(x1, SIZE(b)),  __jjj__ = min(x2, SIZE(b) - 1); __iii__ <= __jjj__; ++__iii__) { t += b[__iii__] + '0'; } return '"' + t + '"'; } template <typename A, typename... C> string to_string(const A (&v), int x1 = 0, int x2 = 1e9, C... coords); int l_v_l_v_l = 0, t_a_b_s = 0; template <typename A, typename B> string to_string(const pair<A, B> &p) { l_v_l_v_l++; string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; l_v_l_v_l--; return res; } template <typename A, typename... C> string to_string(const A (&v), int x1, int x2, C... coords) { int rnk = rank<A>::value; string tab(t_a_b_s, ' '); string res = ""; bool first = true; if (l_v_l_v_l == 0) res += "\n"; res += tab + "["; x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v)); auto l = begin(v); advance(l, x1); auto r = l; advance(r, (x2 - x1) + (x2 < SIZE(v))); for (auto e = l; e != r; e = next(e)) { if (!first) { res += ", "; } first = false; l_v_l_v_l++; if (e != l) { if (rnk > 1) { res += "\n"; t_a_b_s = l_v_l_v_l; }; } else { t_a_b_s = 0; } res += to_string(*e, coords...); l_v_l_v_l--; } res += "]"; if (l_v_l_v_l == 0) res += "\n"; return res; } void dbgm() {;} template<typename Heads, typename... Tails> void dbgm(Heads H, Tails... T) { cerr << to_string(H) << " | "; dbgm(T...); }
#define debm(...) cerr << "[" << #__VA_ARGS__ << "]: "; dbgm(__VA_ARGS__); cerr << endl

const int N = 2e5 + 6;
vi v = {3, 5, 7, 9, 11, 13, 17, 19, 23, 27, 29};
signed main()
{

    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("error.txt", "w", stderr);
#endif
    int _;
    cin >> _;
    while (_--) {
        int n;
        cin >> n;
        int x = n / 2;
        if (n & 1) {
            for (int i = 0; i < size(v); i++) {
                int p = v[i];
                if ((n - p - 1) % p) {
                    cout << p << " " << n - p - 1 << " " << 1 << '\n';
                    break;
                }
            }
        }
        else {
            cout << x << " " << x - 1 << " " << 1 << '\n';
        }
    }
    return 0;

}
