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: # cook your dish here
  # cook your dish here
  #-------------------imports-------#
  import sys
  import io,os
  import string
  from io import BytesIO, IOBase
  from os import path
  from collections import Counter,OrderedDict
  import itertools
  
  from itertools import chain, combinations
  import math
  import random
  import bisect
  from functools import reduce
  from sys import stdin,stdout
  from math import gcd,floor,sqrt,log
  from collections import defaultdict as dd
  from bisect import bisect_left as bl,bisect_right as br
  
  
  #------------------ONLINE JUDGE OR NOT-------#
  
  if(path.exists('input.txt')):
      sys.stdin = open("input.txt","r")
      sys.stdout = open("output.txt","w")
  else:
      input = sys.stdin.readline
      # input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  
  
  
  sys.setrecursionlimit(100000000)
  
  
  
  #---------------------pre made and input functions-------#
  simp    =lambda: int(input())
  strng  =lambda: input().strip()
  jn     =lambda x,l: x.join(map(str,l))
  strl   =lambda: list(input().strip())
  mul    =lambda: map(int,input().strip().split())
  mulf   =lambda: map(float,input().strip().split())
  seq    =lambda: list(map(int,input().strip().split()))
  ceil   =lambda x: int(x) if(x==int(x)) else int(x)+1
  ceildiv=lambda x,d: x//d if(x%d==0) else x//d+1
  flush  =lambda: stdout.flush()
  stdstr =lambda: stdin.readline()
  stdint =lambda: int(stdin.readline())
  stdpr  =lambda x: stdout.write(str(x))
  
  
  #------------functions---------------------
  
  #---------------------input-------
  
  n = simp()
  
  
  def if_its_binaryform_palindrome(n):
  
      res = 0
  
      k = n
      while k > 0:
          #to finish the job
          res = (res << 1) | (k & 1)
          k = k >> 1  
  
      return n == res
  
  
  
  
  
  
  def solve(d):
  
      a,b,c = d[0],d[1],d[2]
  
      if a==b:
          if c&1==0:
              return 'YES'
          return 'NO'
      elif b==c:
          if a&1==0:
              return 'YES'
          return 'NO'
      elif a==c:
          if b&1==0:
              return 'YES'
          return 'NO'    
  
  
      if a+b==c or b+c==a or a+c==b:
          return 'YES'
      return 'NO'    
  
  for i in range(n):
      # d = seq()
      a = seq()
      # d = seq()
      print(solve(a))
      # r = strng()
      # t = strng()
      # d = strng()
      # t = seq()
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  def seive(n):
      """ preprocessing  """
   
      prime = [1 for i in range(n + 1)]
      p = 2
      while (p * p <= n):
           
          # this is prime number.
          if (prime[p] == 1):
               
              # Update all multiples of p in array
              for i in range(p ** 2, n + 1, p):
                  prime[i] = 0
                  
          p += 1
      prime[0]= 0
      prime[1]= 0
      
      # print(prime)
      for i in range(1,n+1):
          prime[i] += prime[i-1]
          
      return prime
  
  def list_of_prime_sieve(n):
  
      #an array of True True
      primes = [True]*(n+1)
      #changing o and 1 to False
      primes[0]==primes[1]==False
  
      #now for every true we find
      for i in range(2,int(n**(1/2))+1):
          if primes[i]==True:
              #we change their multiples to False
              for j in  range(i*i,n+1,i):
                  primes[j]=False
  
  
      list_primes = []
      for i in range(2,len(primes)):
          if primes[i]==True:
              list_primes.append(i)
  
      return list_primes
  
  def count_primes(a):
      return len(list_of_prime_sieve(a))    
  
  
  def gcdoftwo(a,b):
  
      #log(min(a,b))
      if a == 0:
          return b
      else:
          return gcdoftwo(b%a,a)
  
  def lcm(a,b): return (a*b//(gcdoftwo(a,b)))        
  
  def gcdofList(A): 
      res = A[0]
      for c in A[1::]:
          res = gcdoftwo(res , c) 
          if res == 1:
              return res
      return res         
  
  def binpow(a,b):
  
      #to calc a^b in log(n) time
  
  
      res = 1
  
  
      while b>0:
  
          if b&1:
              res = res*a
          a = a*a
          b>>=1
      return res        
  
  def fibo_in_logn(n):
      if n == 0:
          return [0, 1]
  
      p = fibo_in_logn(n >> 1)
      c = p.first * (2 * p.second - p.first)
      d = p.first * p.first + p.second * p.second
      if n & 1:
          return {d, c + d}
      else:
          return {c, d}
  
  
  
  
  
  def x_n_mod_m(a,b,m):
  
      #to calculate x^n mod m
  
      a  = a%m 
  
      res = 1
      while b>0:
          if b&1:
              res = res*a%m 
          a = a*a%m
          b>>=1
  
      return res
  
  
  
  
  def change_array(count,t):
  
      emp = []
      for i in range(len(t)):
          emp.append(count[t[i]])
      return emp    
  
  
  
  
  # prime = []
  
  def simpleSieve(limit):
  
      mark = [True for i in range(limit + 1)]
      p = 2
      while (p * p <= limit):
          if (mark[p] == True):
              for i in range(p * p, limit + 1, p):
                  mark[i] = False
          p += 1
      for p in range(2, limit):
          if mark[p]:
              prime.append(p)
              # print(p,end = " ")
      return prime        
  
  def segmentedSieve(n):
      limit = int(math.floor(math.sqrt(n)) + 1)
      t = []
      t += simpleSieve(limit)
  
      low = limit
      high = limit * 2
      while low < n:
          if high >= n:
              high = n
  
          mark = [True for i in range(limit + 1)]
  
          for i in range(len(prime)):
  
              loLim = int(math.floor(low / prime[i]) *
                                          prime[i])
              if loLim < low:
                  loLim += prime[i]
              for j in range(loLim, high, prime[i]):
                  mark[j - low] = False
  
          # t = []        
          for i in range(low, high):
              if mark[i - low]:
                  t.append(i)
                  # print(i, end = " ")
  
          low = low + limit
          high = high + limit
  
      return t
  
  
  
  
  mod=1000000007
  
  def occurrences(string, sub):
      count = start = 0
      while True:
          start = string.find(sub, start) + 1
          if start > 0:
              count+=1
          else:
              return count
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
      
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  