Codeforces Round #657 (Div. 2) B. Dubious Cyrpto（暴力/数学）

Pasha loves to send strictly positive integers to his friends. Pasha cares about security, therefore when he wants to send an integer nn , he encrypts it in the following way: he picks three integers aa , bb and cc such that l≤a,b,c≤rl≤a,b,c≤r , and then he computes the encrypted value m=n⋅a+b−cm=n⋅a+b−c .

Unfortunately, an adversary intercepted the values ll , rr and mm . Is it possible to recover the original values of aa , bb and cc from this information? More formally, you are asked to find any values of aa , bb and cc such that

• aa , bb and cc are integers,
• l≤a,b,c≤rl≤a,b,c≤r ,
• there exists a strictly positive integer nn , such that n⋅a+b−c=mn⋅a+b−c=m .

Input

The first line contains the only integer tt (1≤t≤201≤t≤20 ) — the number of test cases. The following tt lines describe one test case each.

Each test case consists of three integers ll , rr and mm (1≤l≤r≤5000001≤l≤r≤500000 , 1≤m≤10101≤m≤1010 ). The numbers are such that the answer to the problem exists.

Output

For each test case output three integers aa , bb and cc such that, l≤a,b,c≤rl≤a,b,c≤r and there exists a strictly positive integer nn such that n⋅a+b−c=mn⋅a+b−c=m . It is guaranteed that there is at least one possible solution, and you can output any possible combination if there are multiple solutions.

Example

Input

Copy

2

4 6 13

2 3 1

Output

Copy

4 6 5

2 2 3

对于表达式n * a + b – c = m, 只有m是确定的，又看到abc范围5e5，就能知道只需要枚举一个就好了，自然选择枚举a。因为b，c都在同一个范围内，所以尽量使得a * n更贴近m，便于构造b和c。因此从l~r枚举a，然后分a * n > m和 a * n < m两种情况，看哪种成立即可（没有要求最优啥的）。

特判一下m小于l的情况此时n * a必然大于m，必须让n取1，a取l才能使得b – c尽可能接近m。

记得开long long！

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        long long m, l, r;//别忘记开long long 
        cin >> l >> r >> m;
        long long a = 0, b = 0, c = 0;
        if(m >= l)
        {
            for(a = l; a <= r; a++)
            {
                long long temp = m / a * a;//n * a < m 
                if(m - temp <= r - l)
                {
                    c = l, b = l + m - temp;
                    break;
                } 
                temp += a;//n * a > m 
                if(temp - m <= r - l)
                {
                    b = l, c = l + temp - m;
                    break;
                }
            }  
        }
        else//n只能为1（贪心 
        {
            a = l;
            b = l;
            c = 2 * l - m;
        }
        cout << a << ' ' << b << ' ' << c << ' ' << endl;
     }
    return 0;
}