description: The only difference between this problem and D2 is that you don't have to provide the way to construct the answer in this problem, but you have to do it in D2.
  
  There's a table of n × m cells (n rows and m columns). The value of n ⋅ m is even.
  
  A domino is a figure that consists of two cells having a common side. It may be horizontal (one of the cells is to the right of the other) or vertical (one of the cells is above the other).
  
  You need to find out whether it is possible to place nm/2 dominoes on the table so that exactly k of them are horizontal and all the other dominoes are vertical. The dominoes cannot overlap and must fill the whole table.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10) — the number of test cases. Then t test cases follow.
  
  Each test case consists of a single line. The line contains three integers n, m, k (1 ≤ n,m ≤ 100, 0 ≤ k ≤ nm/2, n ⋅ m is even) — the number of rows, columns and horizontal dominoes, respectively.
  
  Output
  
  For each test case output "YES", if it is possible to place dominoes in the desired way, or "NO" otherwise.
  
  You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
  
  Example
  
  Input
  
  
  8
  4 4 2
  2 3 0
  3 2 3
  1 2 0
  2 4 2
  5 2 2
  2 17 16
  2 1 1
  
  
  Output
  
  
  YES
  YES
  YES
  NO
  YES
  NO
  YES
  NO
solution: #include <bits/stdc++.h>
  using namespace std;
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << ' ' << H;
    dbg_out(T...);
  }
  pair<int, int> operator*(const pair<int, int> &p, const pair<int, int> &x) {
    pair<int, int> np = make_pair(p.first * x.first, p.second * x.second);
    return np;
  }
  pair<long long, long long> operator%(const pair<long long, long long> &p,
                                       const pair<long long, long long> &x) {
    pair<long long, long long> np =
        make_pair(p.first % x.first, p.second % x.second);
    return np;
  }
  pair<long long, long long> operator*(const pair<long long, long long> &p,
                                       const pair<int, int> &x) {
    pair<long long, long long> np =
        make_pair(p.first * x.first, p.second * x.second);
    return np;
  }
  pair<long long, long long> operator*(const pair<long long, long long> &p,
                                       const pair<long long, long long> &x) {
    pair<long long, long long> np =
        make_pair(p.first * x.first, p.second * x.second);
    return np;
  }
  pair<long long, long long> operator+(const pair<long long, long long> &p,
                                       const pair<long long, long long> &x) {
    pair<long long, long long> np =
        make_pair(p.first + x.first, p.second + x.second);
    return np;
  }
  pair<long long, long long> operator-(const pair<long long, long long> &p,
                                       const pair<long long, long long> &x) {
    pair<long long, long long> np =
        make_pair(p.first - x.first, p.second - x.second);
    return np;
  }
  pair<int, int> operator+(const pair<int, int> &p, const pair<int, int> &x) {
    pair<int, int> np = make_pair(p.first + x.first, p.second + x.second);
    return np;
  }
  pair<int, int> operator-(const pair<int, int> &p, const pair<int, int> &x) {
    pair<int, int> np = make_pair(p.first - x.first, p.second - x.second);
    return np;
  }
  vector<int> prime;
  bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
    return (a.first > b.first);
  }
  bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
    return a.second > b.second;
  }
  bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
    return (a.second < b.second);
  }
  long long powerm(long long a, long long k) {
    if (!k) {
      return 1;
    }
    long long b = powerm(a, k / 2);
    b = b * b % 998244353;
    if (k % 2) {
      return a * b % 998244353;
    } else {
      return b;
    }
  }
  long long power(long long a, long long b) {
    if (b == 1) return a;
    if (b == 0) return 1;
    long long m1 = power(a, b / 2);
    if (b % 2) return m1 * m1 * a;
    return m1 * m1;
  }
  void PFactor(int x) {
    prime.clear();
    for (int i = 2; i <= x / i; i++)
      if (x % i == 0) {
        while (x % i == 0) x /= i, prime.push_back(i);
      }
    if (x > 1) prime.push_back(x);
  }
  bool isprime(long long a) {
    if (a <= 1) return false;
    if (a == 2 || a == 3) return true;
    if (a % 2 == 0 || a % 3 == 0) return false;
    for (long long i = 5; i * i <= a; i = i + 6) {
      if (a % i == 0 || a % (i + 2) == 0) return false;
    }
    return true;
  }
  long long read() {
    long long a = 0;
    char b = 1, c;
    do
      if ((c = getchar()) == 45) b = -1;
    while (c < 48 || c > 57);
    do a = (a << 3) + (a << 1) + (c & 15);
    while ((c = getchar()) > 47 && c < 58);
    return a * b;
  }
  void write(long long x, char c) {
    if (x < 0) putchar(45), x = -x;
    char a[20], s = 0;
    do a[++s] = x % 10 | 48;
    while (x /= 10);
    do putchar(a[s]);
    while (--s);
    putchar(c);
  }
  const int N = 2e6 + 20;
  vector<int> g[N];
  int n = 0, m = 0, k = 0;
  int a = 0, b = 0, c = 0;
  string str;
  int d[N] = {0};
  int main() {
    int T;
    cin >> T;
    while (T--) {
      cin >> n >> m >> k;
      a = b = c = 0;
      int sum = n * m;
      if (n % 2 == 1) {
        sum -= m;
        a += m / 2;
      } else if (m % 2 == 1) {
        sum -= n;
        b += n / 2;
      }
      if (k < a || (k - a) / 2 > sum / 4 || (k - a) % 2 != 0) {
        cout << "NO\n";
      } else
        cout << "YES\n";
    }
  }
  