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:
t=int(input())
for _ in range(t):
    x,y,z=map(int,input().split())
    l=[x,y,z]
    if l.count(max(l))<2:
        print('NO')
    else:
        print('YES')
        mi=min(l)
        mx=max(l)
        idx=l.index(mi)
        if idx==0:
            print(mi,mi,mx)
        elif idx==1:
            print(mx,mi,mi)
        else:
            print(mi,mx,mi)
        









'''
import math 
def pf(n):
    a=[1]
    while n % 2 == 0: 
        a.append(2)
        n = n / 2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0:
            a.append(i) 
            n = n / i 
    if n > 2: 
        a.append(n)
    return sorted(a)


t=int(input())
for _ in range(t):
    n=int(input())
    if n%2==0:
        print(n//2,n//2)
    else:
        sol=pf(n)
        if sol[-1]==n:
            out=sol[-2]
        else:
            out=sol[-1]
        
        print(int(out),int(n-out))
'''
'''
t=int(input())
for _ in range(1,t+1):
    n=int(input())
    l=list(map(int,input().split()))
    s=['A','B','C','D']
    m=l[0]
    ml=[0]
    mil=[0]
    mi=l[0]
    cm=0
    cmi=0
    for i in range(n):
        if l[i]>m:
            m=l[i]
            cm+=1
            #print(cm)
            ml.append(cm)
        else:
            cm=0
            m=l[i]
            #print(cm)
        if l[i]<mi:
            mi=l[i]
            cmi+=1
            mil.append(cmi)
        else:
            cmi=0
            mi=l[i]
    ma=max(ml)+1
    mi=max(mil)+1
    ma=(ma//4)-1
    mi=(mi//4)-1
    out=0
    for i in (ma,mi):
        if i>0:
            out+=i
    print('Case #{}: {}'.format(_,out))
'''
'''
def f(k, arr):
    if k<0:
        return 0
    if k==0:
        return 1
    for j in arr:
        t=f(k-j,arr)
        if t:
            return k
    return 0
def unboundedKnapsack(k, arr):
    if k<0:
        return 0
    if k==0:
        return 1
    for i in range(k,-1,-1):
        o=f(i,arr)
        #print(i,o)
        if o:
            return i
t=int(input())
out=[]
for _ in range(t):
    n,k=map(int,input().split())
    arr=list(map(int,input().split()))
    out.append(unboundedKnapsack(k,arr))
for i in out:
    print(i)
'''
'''
n=17048
k=99
x=1
#761 99 1
#17048 14319 1
#n,k,x=map(int,input().split())
def countArray(n, k, x):
    # Return the number of ways to fill in the array.
    dp=[[0]*(k+1) for i in range(n+1)]
    dp[1][1]=1
    for i in range(2,n+1):
        dp[i][1]=(dp[i-1][2]*(k-1))%1000000007
        dp[i][2]=(dp[i-1][1] + (dp[i-1][2]*(k-2)))%1000000007
    if(x==1):
        return dp[n][1]
    else:
        return dp[n][2]

print(countArray(n,k,x))
'''