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.
2
1 2 3
2 2 4
 1
1 98 99
 3
1 1 1
2 1 3
5 6 7
 6
2 2 4
1 1 2
3 5 6
1 2 3
6 7 8
10 10 20
 4
6 1 5
2 5 2
2 4 2
7 101 108
 1
100000 100000 100000
 1
12344 1 1
 68
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
1 1 2
 2
1 2 3
4 5 6

YES
YES
 YES
 NO
YES
NO
 YES
YES
NO
YES
NO
YES
 YES
NO
YES
YES
 YES
 YES
 YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
 YES
NO

#include <bits/stdc++.h>
using namespace std;
template <class T>
using vc = vector<T>;
template <typename T1, typename T2>
pair<T1, T2>& operator+=(pair<T1, T2>& p1, const pair<T1, T2>& p2) {
  p1.first += p2.first, p1.second += p2.second;
  return p1;
}
template <typename T1, typename T2>
pair<T1, T2>& operator-=(pair<T1, T2>& p1, const pair<T1, T2>& p2) {
  p1.first -= p2.first, p1.second -= p2.second;
  return p1;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> p1, const pair<T1, T2>& p2) {
  return p1 += p2;
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> p1, const pair<T1, T2>& p2) {
  return p1 -= p2;
}
template <typename T>
T operator*(const pair<T, T>& p1, const pair<T, T>& p2) {
  return p1.first * p2.first + p1.second * p2.second;
}
template <typename T>
T operator^(const pair<T, T>& p1, const pair<T, T>& p2) {
  return p1.first * p2.second - p2.first * p1.second;
}
template <typename T>
pair<T, T>& operator*=(pair<T, T>& p1, const T& p2) {
  p1.first *= p2;
  p1.second *= p2;
  return p1;
}
template <typename T>
pair<T, T> operator*(pair<T, T> p1, const T& p2) {
  return p1 *= p2;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << p.first << ' ' << p.second;
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vc<T>& p) {
  for (auto& v : p) os << (v) << ' ';
  return os;
}
template <typename T>
istream& operator>>(istream& is, vc<T>& p) {
  for (auto& v : p) is >> v;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vc<vc<T>>& p) {
  for (auto& v : p) os << v << '\n';
  return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const map<T1, T2>& p) {
  for (auto& v : p)
    os << '\n' << setw(15) << v.first << " : " << v.second << ' ';
  return os;
}
template <typename T>
vc<T>& operator+=(vc<T>& p, const vc<T>& b) {
  for (auto& v : b) p.emplace_back(v);
  return p;
}
template <typename T>
vc<T> operator+(vc<T> p, const vc<T>& b) {
  return p += b;
}
template <typename T>
vc<T>& operator+=(vc<T>& p, const T& b) {
  for (auto& v : p) v += b;
  return p;
}
long long gcd(long long p, long long b) { return b ? gcd(b, p % b) : p; }
long long lcd(long long p, long long b) { return p * b / gcd(p, b); }
long long qmi(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % p;
      b >>= 1;
    }
    a = (a * a) % p;
  }
  return res;
}
void dbg_out() { cerr << "\b\b )" << '\n'; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << H << ", ";
  dbg_out(T...);
}
const int P = 13331;
const double eps = 1e-10;
const int N = 2e5 + 10, M = 2 * N, MOD = 1e9 + 7, INF = 0x3f3f3f3f,
          mod = 998244353;
long long n, m, k;
void solve(int caseT) {
  int a[3];
  int b = 0, c = 0;
  for (int i = 0; i < 3; i++) {
    cin >> a[i];
    if (a[i] & 1)
      b++;
    else
      c++;
  }
  if (a[0] == a[1] == a[2]) {
    cout << "YES" << '\n';
    return;
  }
  if (b == 2 && c == 1)
    cout << "YES" << '\n';
  else if (c == 3)
    cout << "YES" << '\n';
  else
    cout << "NO" << '\n';
}
void init() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  init();
  int T;
  T = 1;
  cin >> T;
  for (int i = 1; i <= T; i++) solve(i);
  return 0;
}
