2016 Al-Baath University Training Camp Contest-1 G

Description

The forces of evil are about to disappear since our hero is now on top on the tower of evil, and all what is left is the most evil, most dangerous monster! The tower has h floors (numbered from 1 to h, bottom to top), each floor has w rooms (numbered from 1 to w, left to right) composing a row. The monster stands in one of the ground floor (floor number 1) rooms (the room number d where 1 ≤ d ≤ w). Our hero stands in the top-left room of the tower. The only way for our hero to kill the evil monster is to throw down one of his power stones diagonally to the right. The stone will keep moving diagonally through the tower's rooms until it hits the right border of the tower, then it will change its direction to move down diagonally to the left until it hits the left border of the tower, then it will change its direction again and so on. This stone stops when it reaches the ground floor, if it stopped in the room of the monster, the monster is dead, otherwise the monster is still alive and the forces of evil will rise again! Help our hero to determine whether the stone will kill the monster or not!

Input

The input consists of several test cases. The first line of the input contains a single integer T, the number of the test cases. Each of the following lines contains a test case and consists of a three space-separated integers h, w and d denoting the height of the tower, the width of the tower and the number of room containing the monster. (1 ≤ d ≤ 109), (2 ≤ h, w ≤ 109).

Output

For each test case print a single line: 'Yes' if the monster will be killed and 'No' otherwise.

Example

input

output

No
Yes
Yes
No
Yes

Note

$\text{[math]}$

In the first test case, the path of the power stone is colored in red. However the dragon is in the room denoted with D, so the monster is still alive and the answer is 'No'.

题意：见图

解法：找规律，这里以2*w-2作为一个循环，循环里面以最高点为对称

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t,w,h,d;
    cin>>t;
    while(t--)
    {
        cin>>h>>w>>d;
        int num=2*w-2;
        int mod=h%num;
        //cout<<h%num<<endl;
        if(mod>w)
        {
            mod=2*w-mod;
        }
        else if(mod==0)
        {
            mod=2;
        }
        if(mod==d)
        {
            cout<<"Yes"<<endl;
        }
        else
        {
            cout<<"No"<<endl;
        }
        //cout<<mod<<endl;
    }
    return 0;
}