description: There are three sticks with integer lengths l_1, l_2 and l_3.
  
  You are asked to break exactly one of them into two pieces in such a way that: 
  
    * both pieces have positive (strictly greater than 0) integer length; 
    * the total length of the pieces is equal to the original length of the stick; 
    * it's possible to construct a rectangle from the resulting four sticks such that each stick is used as exactly one of its sides. 
  
  
  
  A square is also considered a rectangle.
  
  Determine if it's possible to do that.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.
  
  The only line of each testcase contains three integers l_1, l_2, l_3 (1 ≤ l_i ≤ 10^8) — the lengths of the sticks.
  
  Output
  
  For each testcase, print "YES" if it's possible to break one of the sticks into two pieces with positive integer length in such a way that it's possible to construct a rectangle from the resulting four sticks. Otherwise, print "NO".
  
  You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as a positive answer).
  
  Example
  
  Input
  
  
  4
  6 1 5
  2 5 2
  2 4 2
  5 5 4
  
  
  Output
  
  
  YES
  NO
  YES
  YES
  
  Note
  
  In the first testcase, the first stick can be broken into parts of length 1 and 5. We can construct a rectangle with opposite sides of length 1 and 5.
  
  In the second testcase, breaking the stick of length 2 can only result in sticks of lengths 1, 1, 2, 5, which can't be made into a rectangle. Breaking the stick of length 5 can produce results 2, 3 or 1, 4 but neither of them can't be put into a rectangle.
  
  In the third testcase, the second stick can be broken into parts of length 2 and 2. The resulting rectangle has opposite sides 2 and 2 (which is a square).
  
  In the fourth testcase, the third stick can be broken into parts of length 2 and 2. The resulting rectangle has opposite sides 2 and 5.
solution: from __future__ import division, print_function
  import math
  import sys
  import os
  from io import BytesIO, IOBase
  #from collections import deque, Counter, OrderedDict, defaultdict
  #import heapq
  #ceil,floor,log,sqrt,factorial,pow,pi,gcd
  #import bisect
  #from bisect import bisect_left,bisect_right
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
  	newlines = 0
  
  	def __init__(self, file):
  		self._fd = file.fileno()
  		self.buffer = BytesIO()
  		self.writable = "x" in file.mode or "r" not in file.mode
  		self.write = self.buffer.write if self.writable else None
  
  	def read(self):
  		while True:
  			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
  			if not b:
  				break
  			ptr = self.buffer.tell()
  			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
  		self.newlines = 0
  		return self.buffer.read()
  
  	def readline(self):
  		while self.newlines == 0:
  			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
  			self.newlines = b.count(b"\n") + (not b)
  			ptr = self.buffer.tell()
  			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
  		self.newlines -= 1
  		return self.buffer.readline()
  
  	def flush(self):
  		if self.writable:
  			os.write(self._fd, self.buffer.getvalue())
  			self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
  	def __init__(self, file):
  		self.buffer = FastIO(file)
  		self.flush = self.buffer.flush
  		self.writable = self.buffer.writable
  		self.write = lambda s: self.buffer.write(s.encode("ascii"))
  		self.read = lambda: self.buffer.read().decode("ascii")
  		self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  def print(*args, **kwargs):
  	"""Prints the values to a stream, or to sys.stdout by default."""
  	sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
  	at_start = True
  	for x in args:
  		if not at_start:
  			file.write(sep)
  		file.write(str(x))
  		at_start = False
  	file.write(kwargs.pop("end", "\n"))
  	if kwargs.pop("flush", False):
  		file.flush()
  
  
  if sys.version_info[0] < 3:
  	sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
  	sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  def I():
      return(int(input()))
  def lint():
      return(list(map(int,input().split())))
  def insr():
      s = input().strip()
      return(list(s[:len(s)]))
  def invr():
      return(map(int,input().split()))
  import itertools
  from sys import maxsize, stdout, stdin,stderr
  mod = int(1e9+7)
  import sys
  # def I(): return int(stdin.readline())
  # def lint(): return [int(x) for x in stdin.readline().split()]
  def S(): return list(map(str,input().strip()))
  def grid(r, c): return [lint() for i in range(r)]
  from collections import defaultdict, Counter, deque
  import math
  import heapq
  from heapq import heappop , heappush
  import bisect
  from math import factorial, inf
  from itertools import groupby
  from itertools import permutations as comb
  
  mod=1000000007
  MOD = 998244353
  def gcd(a,b): 
      while b:
          a %= b
          tmp = a
          a = b
          b = tmp
      
      return a
   
  def lcm(a,b): 
      return a  // gcd(a, b) * b
   
  def check_prime(n):
      for i in range(2, int(n ** (1 / 2)) + 1):
          if not n % i:
              return False
      return True
  
  # def nCr(n, r):
   
  #     return (fact(n) // (fact(r)
  #                 * fact(n - r)))
   
  # Returns factorial of n
  # def fact(n):
   
  #     res = 1
       
  #     for i in range(2, n+1):
  #         res = res * i
           
  #     return res
  def primefactors(n):
      num=0
      
      while n % 2 == 0:
          num+=1
          n = n / 2
      
      for i in range(3,int(math.sqrt(n))+1,2):
           
      
          while n % i== 0:
              num+=1
              n = n // i
                 
      
      if n > 2:
          num+=1
      return num
  
  def ask(a,b,c):
      # print('? 1 {}'.format(a),flush=True)
      print(c,a,b,flush=True)
      n=I()
      
      return n
  def linear_sieve(n):
      is_composite=[False]*n
      prime=[]
      for i in range(2,n):
          if not is_composite[i]:
              prime.append(i)
          for j in prime:
              is_composite[i*j]=True
              if i%prime==0:
                  break
      return prime
   
  
  def primeFactors(n):
      l=[]
      
      while n % 2 == 0:
          l.append(2)
          n = n // 2
      for i in range(3,int(math.sqrt(n))+1,2):
           
          
          while n % i== 0:
              l.append(i)
              n = n // i
  
      if n > 2:
          l.append(n)
      return l
  
  def binpow(a,b,c):
      res=1
      a=a%c
      while b>0:
          if b&1:
              res=(res*a)%c
  
          b=b>>1
          a = (a*a)%c
  
      return res
  
  
      
  
  #  Sieve
  d=[]
  primes=[]
  prim=[0]*(10**5+1)
  def sieve(n):
      
      for i in range(n):
          d.append(i)
  
      for i in range(2,n):
  
          if d[i]==i:
              prim[i]=1
              primes.append(i)
          j=0
          while j<len(primes) and primes[j]<=d[i] and primes[j]*i<n:
          
              d[i * primes[j]] = primes[j]
              j+=1
  
          
  def primeFactors(n):
      factors=[]
      while n!=1:
          factors.append(d[n])
          n//=d[n]
      
      return factors
  
  
  
  t = 1
  t = I()
  for _ in range(t):
      s = lint()
      s.sort()
      a,b,c=s
      if(a+b==c):
          print("YES")
      elif a==b and c%2==0:
          print("YES")
      elif b==c and a%2==0:
          print("YES")
      else:
          print("NO")
      
          
  
  
  
  
  
          
      
   
  
  