B. Make AP【模拟】

Problem - B - Codeforces

Polycarp has 33 positive integers aa, bb and cc. He can perform the following operation exactly once.

• Choose a positive integer mm and multiply exactly one of the integers aa, bb or cc by mm.

Can Polycarp make it so that after performing the operation, the sequence of three numbers aa, bb, cc (in this order) forms an arithmetic progression? Note that you cannot change the order of aa, bb and cc.

Formally, a sequence x1,x2,…,xnx1,x2,…,xn is called an arithmetic progression (AP) if there exists a number dd (called "common difference") such that xi+1=xi+dxi+1=xi+d for all ii from 11 to n−1n−1. In this problem, n=3n=3.

For example, the following sequences are AP: [5,10,15][5,10,15], [3,2,1][3,2,1], [1,1,1][1,1,1], and [13,10,7][13,10,7]. The following sequences are not AP: [1,2,4][1,2,4], [0,1,0][0,1,0] and [1,3,2][1,3,2].

You need to answer tt independent test cases.

Input

The first line contains the number tt (1≤t≤1041≤t≤104) — the number of test cases.

Each of the following tt lines contains 33 integers aa, bb, cc (1≤a,b,c≤1081≤a,b,c≤108).

Output

For each test case print "YES" (without quotes) if Polycarp can choose a positive integer mm and multiply exactly one of the integers aa, bb or cc by mm to make [a,b,c][a,b,c] be an arithmetic progression. Print "NO" (without quotes) otherwise.

You can print YES and NO in any (upper or lower) case (for example, the strings yEs, yes, Yes and YES will be recognized as a positive answer).

Example

input

Copy

11
10 5 30
30 5 10
1 2 3
1 6 3
2 6 3
1 1 1
1 1 2
1 1 3
1 100000000 1
2 1 1
1 2 2

output

Copy

YES
YES
YES
YES
NO
YES
NO
YES
YES
NO
YES

Note

In the first and second test cases, you can choose the number m=4m=4 and multiply the second number (b=5b=5) by 44.

In the first test case the resulting sequence will be [10,20,30][10,20,30]. This is an AP with a difference d=10d=10.

In the second test case the resulting sequence will be [30,20,10][30,20,10]. This is an AP with a difference d=−10d=−10.

In the third test case, you can choose m=1m=1 and multiply any number by 11. The resulting sequence will be [1,2,3][1,2,3]. This is an AP with a difference d=1d=1.

In the fourth test case, you can choose m=9m=9 and multiply the first number (a=1a=1) by 99. The resulting sequence will be [9,6,3][9,6,3]. This is an AP with a difference d=−3d=−3.

In the fifth test case, it is impossible to make an AP.

题意：给你三个数a，b，c 判断某个数乘以一个正整数m，能否使该三个数为等差数列。

 

 咳咳，这图。。

#include<iostream>
using namespace std;
int t;
int a,b,c;
int main()
{
    scanf("%d", &t);
    while(t--)
    {
        int flag=0;
        scanf("%d %d %d",&a,&b,&c);
        if(b-a==c-b)flag=1;
        else
        {
            int x=(b-(c-b));
            int y=(c-((c-a)/2));
            int z=(b+(b-a));
            if((x%a==0)&&x>0)
            {
                a*=x/a;
                if(b-a==c-b)
                  flag=1;
                
            }
            else if(y%b==0&&y>0)
            {
                b*=y/b;
               if(b-a==c-b)
                flag=1;
                
            }
            else if(z%c==0&&z>0)
            {
                c*=z/c;
               if(b-a==c-b)
                flag=1;
              
            }
        }
        if(flag)cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
}