A. Three Pairwise Maximums

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given three positive (i.e. strictly greater than zero) integers xx, yy and zz.

Your task is to find positive integers aa, bb and cc such that x=max(a,b)x=max(a,b), y=max(a,c)y=max(a,c) and z=max(b,c)z=max(b,c), or determine that it is impossible to find such aa, bb and cc.

You have to answer tt independent test cases. Print required aa, bb and cc in any (arbitrary) order.

Input

The first line of the input contains one integer tt (1≤t≤2⋅1041≤t≤2⋅104) — the number of test cases. Then tt test cases follow.

The only line of the test case contains three integers xx, yy, and zz (1≤x,y,z≤1091≤x,y,z≤109).

Output

For each test case, print the answer:

• "NO" in the only line of the output if a solution doesn't exist;
• or "YES" in the first line and any valid triple of positive integers aa, bb and cc (1≤a,b,c≤1091≤a,b,c≤109) in the second line. You can print aa, bb and cc in any order.

Example

input

Copy

5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000

output

Copy

YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000

解题说明：此题是一道模拟题，根据题目意思进行x,y,z三个值得比较，然后求解，a,b,c就从x,y,z中选择即可。

#include

int main() {
	int t, s;
	int x, y, z;
	scanf("%d", &t);
	for (s = 0; s < t; s++) 
	{
		scanf("%d %d %d", &x, &y, &z);
		if (x>y && x == z) 
		{
			printf("YES\n%d %d %d\n", x, y, y);
		}
		else if (y>x && y == z) 
		{
			printf("YES\n%d %d %d\n", y, x, x);
		}
		else if (x == y && y > z) 
		{
			printf("YES\n%d %d %d\n", x, z, z);
		}
		else if (x == y && x == z) 
		{
			printf("YES\n%d %d %d\n", x, x, x);
		}
		else 
		{
			printf("NO\n");
		}
	}
	return 0;
}