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

//~ author      : Sumit Prajapati
#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;


//---------------------Macros----------------------------------------------------------------------------------
#define ull                     unsigned long long
#define ll                      long long
#define int                     long long
#define pii                     pair<int, int>
#define pll                     pair<ll, ll>
#define pb                      push_back
#define mk                      make_pair
#define ff                      first
#define ss                      second
#define SZ(x)                   ((int)x.size())
#define set_bits                __builtin_popcountll
#define all(a)                  a.begin(),a.end()
#define vec                     vector
#define trav(x,v)               for(auto &x:v)
#define rep(i,n)                for(int i=0;i<n;i++)
#define repe(i,n)               for(int i=1;i<=n;i++)
#define FOR(i,a,b)              for(int i=a;i<=b;i++)
#define curtime                 chrono::high_resolution_clock::now()
#define timedif(start,end)      chrono::duration_cast<chrono::nanoseconds>(end - start).count()
#define myshuffle(a,n)          FOR(i,1,n-1) swap(a[i], a[rand() % (i + 1)])
#define shuffle(a)              shuffle(a.begin(), a.end(), rng)
#define mtrand(a,b)             uniform_int_distribution<int>(a, b)(rng)
#define ordered_set(T)          tree< T ,  null_type ,  less<T> ,  rb_tree_tag ,  tree_order_statistics_node_update >
#define myunique(v)             sort( vec.begin(), vec.end() );vec.erase( unique( vec.begin(), vec.end() ), vec.end() )
// ------------------------------------------------------------------------------------------------------------


// -----------------------------Debugging----------------------------------------------------------------------
template<class Fun> class y_combinator_result {
    Fun fun_;
    public:
        template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
        template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename TT, typename... UU> void dbg_out(TT H, UU... T) { cerr << ' ' << H; dbg_out(T...); }
#ifndef ONLINE_JUDGE
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif
// -----------------------------------------------------------------------------------------------------------



// ------------------------------Global Variables-------------------------------------------------------------
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MD = 1e9 + 7;
const int MDL = 998244353;
const int INF = 1e9;
const int MX = 1e5 + 5;
// -----------------------------------------------------------------------------------------------------------

// --------------------------------Let's Go!------------------------------------------------------------------




struct Testcase{
        
    
    void solve() {

        int a[3];
        int n=3;
        rep(i,n)
            cin>>a[i];
        sort(a,a+n);
        if(a[2]==a[1]){
            if(a[0]%2==0)
                cout<<"YES\n";
            else
                cout<<"NO\n";
        }
        else{
            if(a[0]+a[1]==a[2]){
                cout<<"YES\n";
            }
            else
                cout<<"NO\n";
        }


    }

};

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    auto time0 = curtime;
    int t = 1;
    cin >> t;
    repe(tt, t) {
        // cout<<"Case #"<<tt<<": ";
        Testcase T;
        T.solve();
    }

    // cerr<<"Execution Time: "<<timedif(time0,curtime)*1e-9<<" sec\n";
    return 0;

}