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

/*
    # Enjoy the journey #
*/
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2", "popcnt")
#include <string>
#include <bits/functexcept.h>
#include <iosfwd>
#include <bits/cxxabi_forced.h>
#include <bits/functional_hash.h>
#pragma push_macro("__SIZEOF_LONG__")
#pragma push_macro("__cplusplus")
#define __SIZEOF_LONG__ __SIZEOF_LONG_LONG__
#define unsigned unsigned long
#define __cplusplus 201102L
#define __builtin_popcountl __builtin_popcountll
#define __builtin_ctzl __builtin_ctzll
#include <bitset>
#pragma pop_macro("__cplusplus")
#pragma pop_macro("__SIZEOF_LONG__")
#undef unsigned
#undef __builtin_popcountl
#undef __builtin_ctzl
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ll long long
#define ld long double
#define pb emplace_back
#define _DEBUG(x) cout<< #x <<" = "<<"\""<< (x) <<"\""<<endl
#define forq(i , a , b) for (int i = (a); i <= (b); ++i)
#define qrof(i , b , a) for (int i = (a); i >= (b); --i)
#define forr(i , b) forq( i , 0 , b - 1 )
#define F first
#define S second
#define IF ->first
#define IS ->second
#define endl '\n'
#define OFF false
#define ON true
#define qqmemset(array , val) memset (array , val , sizeof(array))
#define ALLV(vect) vect.begin() , vect.end()
#define mid ((st + en) / 2)
#define mid1 ((2 * st + en) / 3)
#define mid2 ((2 * en + st) / 3)
#define lef 2 * Node
#define rig lef + 1
mt19937 rng( chrono::steady_clock::now().time_since_epoch().count() );
#define Ran(a, b) rng() % ( (b) - (a) + 1 ) + (a)
#define Debug(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
/*
template<
typename Key, // Key type
typename Mapped, // Mapped-policy
typename Cmp_Fn = std::less<Key>, // Key comparison functor
typename Tag = rb_tree_tag, // Specifies which underlying data structure to use
template<
typename Const_Node_Iterator,
typename Node_Iterator,
typename Cmp_Fn_,
typename Allocator_>
class Node_Update = null_node_update, // A policy for updating node invariants
typename Allocator = std::allocator<char> > // An allocator type
class tree;
*/
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {cerr << *it << " = " << a << endl; err(++it, args...);}

ll R = 7 + 1e9 , R1 = 19491001 , R2 = 236 , NUMTESTCASE ;
const ll NN = 10 + 1e6 , Inverse2 = 500000004 ;
const double pi = acos(-1.0) ;
int di [8] = {1 , 0 , -1 , 0  , 1 , -1 , 1  , -1 } , dj [8] = {0 , 1 , 0  , -1 , 1 , -1 , -1 , 1  } ;

int main()
{
    //ios_base::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL) ;
    for (cin >> NUMTESTCASE ; NUMTESTCASE ; NUMTESTCASE --) {
        ll a , b , c ;
        cin >> a >> b >> c;
        bool Ok = false;
        if (a == b) {
            Ok = (c % 2 == 0) ;
        }
        if (a == c)
            Ok = (b % 2 == 0) ;
        if (b == c)
            Ok = (a % 2 == 0) ;
        cout << (Ok  ? "Yes" : "No") << endl ;
    }
    return 0;
}
