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>
#include<sstream>
using namespace std;
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
const long long INF=1e18;
const int inf = 1e9;
const long long mod = 1e9+7;
const int N = 1e6+8;
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define T() int tt; cin>>tt; while(tt--)
#define ll long long
#define prec cout<<fixed<<setprecision(15);
#define ld long double  
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pipi pair<int, pair<int, int>>
#define vii vector<int>
#define F first
#define S second
#define fill(v,p) memset(v, p, sizeof(v))
#define print(v) for(auto &z : v)cout<<z<<" ";cout<<'\n';
#define all(v) v.begin(), v.end()
#define PI 3.141592653589793
#define rank asdf
int main()
{
    T(){
        ll n;
        cin>>n;
        int t1 = 1;
        int t2 = n-1;
        if(t2 % 2 != 0){
            int p1 = t2/2, p2 = t2 - p1;
            cout<<p1<<" "<<p2<<" "<<1<<'\n';continue;
        }
        else{
            t2--;
            int p1 = t2/2 + 1, p2 = t2-p1;
            cout<<p1<<" "<<p2<<" "<<2<<'\n';
            continue;
        }
        bool f = 0;
            for(int i = 2; i*i <= n; i++){
                if(n % i == 0 ){
                    int v = n/i - 1, a1, a2;
                    if(v % 2 == 0){
                        a1 = v/2+1, a2 = v - a1;
                    }
                    else{
                        a1 = v/2, a2 = v - a1;
                    }
                    a1*=i, a2*=i;
                    int a3 = i;
                    if(a1 == a2 || a2 == a3 || a3 == a1)continue;
                    if(__gcd(a1, a2) != i)continue;
                    cout<<a1<<" "<<a2<<" "<<i<<'\n';
                    f = 1;
                    break;
                }
            }
        if(f)continue;
        int a1 = 1;
        int a2 = (n-1)/2 -1, a3 = (n-1)/2 + 1;
        cout<<a2<<" "<<a3<<" "<<a1<<'\n';
    }
}