human
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.
The above is tricky. Write me a correct solution in PYTHON3.

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Give me a Python code to determine if it's possible to do that?
assistant
#Here is Python code to determine if it's possible to do that.

import os
import sys
from io import BytesIO, IOBase
import math
 
#--------------------fastio--------------------

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")

if(os.path.exists('input.txt')):
    sys.stdin = open('input.txt','r') ; sys.stdout = open('output.txt','w')
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
    input = lambda: sys.stdin.readline().rstrip("\r\n")

for t in range(int(input())):
    inp = list(map(int, input().split()))
    inp.sort(reverse=True)
    if(inp[0] == (inp[1]) + inp[2]):
       ...