Problem: You are given three positive (i.e. strictly greater than zero) integers x, y and z.

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

You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).

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 a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order. 

Example

Input


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


Output


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

Solution (in PYTHON3):

Solution:
import sys
import bisect as bi
import math
from collections import defaultdict as dd
input=sys.stdin.readline
##sys.setrecursionlimit(10**7)
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
for _ in range(inin()):
    a,b,c=cin()
    if((a==b and c<=a)):
        print("YES")
        print(a,c,c)
    elif((b==c and a<=c)):
        print("YES")
        print(a,a,b)
    elif(c==a and b<=c):
        print("YES")
        print(b,a,b)
    else:
        print("NO")
        

    
    
    
            
    

    
####    s=input()
####    def minCut( s: str) -> int:
####    
####        n=len(s)
####        dp=[[-1]*(n+2) for i in range(n+2)]
####        def pp(s,i,j,dp,mn):
####            if(i>=j or s[i:j+1]==s[i:j+1:-1]):return 0
####            if(dp[i][j]!=-1):return dp[i][j]
####            for k in range(i,j):
####                if(dp[i][k]!=-1):return dp[i][k]
####                else:
####                    dp[i][k]=pp(s,i,k,dp,mn)
####                    return dp[i][k]
####                if(dp[k+1][j]!=-1):return dp[k+1][j]
####                else:
####                    dp[i][k]=pp(s,k+1,j,dp,mn)
####                    return dp[i][k]
####                temp=dp[i][k] + dp[i][k] +1
####                if(temp<mn):mn=tmp
####            dp[0][n]=mn
####            return dp[0][n]
####        x=pp(s,0,n,dp,999999999999999)
####        for i in dp:print(i)
####        return x
####    
####    print(minCut(s))
##dp=[[0 if (i==j) else -1 for j in range(n+1)] for i in range(n+1)]
##for i in range(1,n):
##    for j in range(
##
##
##
##
##        
##    
##    
####n=m=0
####s=''
####t=''
####dp=[]
####def solve(inds,indt,k,cont):
####    ans=-999999999999999
####    print(dp)
####    if(k<0):return 0
####    elif(inds>=n and indt>=m):return 0
####    elif(dp[inds][indt][k][cont]!=-1):return dp[inds][indt][k][cont]
####    else:
####        if(indt<m):ans=max(ans,solve(inds,indt+1,k,0))
####        if(inds<n):ans=max(ans,solve(inds+1,indt,k,0))
####        if(s[inds]==t[indt]):
####            ans=max(ans,solve(inds+1,indt+1,k-1,1)+1)
####            if(cont):ans=max(ans,solve(inds+1,indt+1,k,1)+1)
####        dp[inds][indt][k][cont]=ans
####        return ans    
##
####    n,m,k=cin()
####    s=sin().strip()
####    t=sin().strip()
####    dp=[[[[-1]*2 for i in range(k)] for i in range(m+1)] for i in range(n+1)]
####    c=0
####    for i in dp:
####        for j in i:
####            for l in j:
####                c+=1
####                print(l,c)
####    print(solve(0,0,k,0))