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: #include <bits/stdc++.h>
  using namespace std;
  long long t;
  int min(long i, long j) {
    if (i >= j) {
      return j;
    }
    return i;
  }
  int max(long i, long j) {
    if (i >= j) {
      return i;
    }
    return j;
  }
  long long Mex(vector<long long> S) {
    long long mex = 0;
    if (S[0] > 0) {
      return 0;
    }
    for (int i = 1; i < S.size(); ++i) {
      if (S[i] - S[i - 1] >= 2) {
        return (S[i - 1] + 1);
      }
      mex = S[i] + 1;
    }
    return mex;
  }
  long long gcd(long long a, long long b) {
    for (long long i = b; i > 1; --i) {
      if (a % i == 0 && b % i == 0) {
        return i;
      }
    }
    return 1;
  }
  int NOD(int a, int b) {
    int div;
    if (a == b) return a;
    int d = a - b;
    if (d < 0) {
      d = -d;
      div = NOD(a, d);
    } else {
      div = NOD(b, d);
    }
    return div;
  }
  long long dist(long long x1, long long y1, long long x2, long long y2) {
    long long x = x1 - x2;
    if (x < 0) x *= -1;
    long long y = y1 - y2;
    if (y < 0) y *= -1;
    return x + y;
  }
  bool del(long long x) {
    for (long long i = 2; i <= sqrt(x); i++) {
      if (x % i == 0) {
        return false;
      }
    }
    return true;
  }
  void find() {
    long long a, b, c;
    cin >> a >> b >> c;
    if (a == b + c || b == a + c || c == a + b) {
      cout << "YES" << endl;
      return;
    }
    if (a == b & c % 2 == 0) {
      cout << "YES" << endl;
      return;
    }
    if (c == b & a % 2 == 0) {
      cout << "YES" << endl;
      return;
    }
    if (a == c & b % 2 == 0) {
      cout << "YES" << endl;
      return;
    }
    cout << "NO" << endl;
  }
  int main() {
    cin >> t;
    while (t--) {
      find();
    }
    return 0;
  }
  