Three Integers

探讨了在三个整数a、b、c中,如何通过增加或减少操作使其形成A≤B≤C且B能被A整除,C能被B整除的序列,并寻找所需的最小操作数。

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You are given three integers a≤b≤ca≤b≤c .
In one move, you can add +1+1 or −1−1 to any of these integers (i.e. increase or decrease any number by one). You can perform such operation any (possibly, zero) number of times, you can even perform this operation several times with one number. Note that you cannot make non-positive numbers using such operations.
You have to perform the minimum number of such operations in order to obtain three integers A≤B≤CA≤B≤C such that BB is divisible by AA and CC is divisible by BB .
You have to answer tt independent test cases.

Input
The first line of the input contains one integer tt (1≤t≤1001≤t≤100 ) — the number of test cases.
The next tt lines describe test cases. Each test case is given on a separate line as three space-separated integers a,ba,b and cc (1≤a≤b≤c≤1041≤a≤b≤c≤104 ).

Output
For each test case, print the answer. In the first line print resres — the minimum number of operations you have to perform to obtain three integers A≤B≤CA≤B≤C such that BB is divisible by AA and CC is divisible by BB . On the second line print any suitable triple A,BA,B and CC .

Example

Input

8
1 2 3
123 321 456
5 10 15
15 18 21
100 100 101
1 22 29
3 19 38
6 30 46

Output

1
1 1 3
102
114 228 456
4
4 8 16
6
18 18 18
1
100 100 100
7
1 22 22
2
1 19 38
8
6 24 48
一种是直接枚举A,B,C
(注意里面两重循环
for(int j = i; j <= 15000; j += i)for(int k = j; k <= 15000; k += j)
不要写++;)
一种是枚举倍数
for(k=1;ijk<=20000;k++)for(k=1;ijk<=20000;k++)。**

#include <bits/stdc++.h>
using namespace std;
int main()
{
 int t;
 cin>>t;
 while(t--)
 {
  int a,b,c,A,B,C,ans=99999999;
  cin>>a>>b>>c;
  for(int i=1;i<=10000;i++)
  {
   for(int j=1;j*i<=20000;j++)
   {
    for(int k=1;i*j*k<=20000;k++)
    {
     int sum=fabs(i-a)+fabs(j*i-b)+fabs(i*j*k-c);
     if(ans>sum)
     {
      ans=sum;
      A=i;
      B=i*j;
      C=i*j*k;
     }
    }
   }
  }
  printf("%d\n",ans);
  printf("%d %d %d\n",A,B,C);
 }
}
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