#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long MOD1 = 998244353;
const long long inf = 1e18;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, -1, 0, 1};
void swap(long long &a, long long &b);
long long m_m(long long a, long long b, long long m);
long long fxp(long long a, long long b, long long m);
long long msum(long long a, long long b, long long m);
long long msub(long long a, long long b, long long m);
long long mpro(long long a, long long b, long long m);
void solve() {
  long long n, m, k, i, j;
  cin >> n >> m >> k;
  long long mnh = 0, mnv = 0;
  if (n & 1) {
    mnh = m / 2;
    if (k < mnh) {
      cout << "NO"
           << "\n";
      return;
    }
    k -= mnh;
    if (k & 1) {
      cout << "NO"
           << "\n";
      return;
    } else {
      cout << "YES"
           << "\n";
      return;
    }
  } else if (m & 1) {
    mnv = n / 2;
    long long avalh = ((n * m) / 2) - mnv;
    if (avalh < k) {
      cout << "NO"
           << "\n";
      return;
    }
    if (k & 1) {
      cout << "NO"
           << "\n";
      return;
    } else {
      cout << "YES"
           << "\n";
      return;
    }
  } else {
    if (k & 1) {
      cout << "NO"
           << "\n";
      return;
    } else {
      cout << "YES"
           << "\n";
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  freopen("Error.txt", "w", stderr);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
long long fxp(long long a, long long b, long long m) {
  if (b == 0) return 1;
  if (b % 2 == 0) return fxp(m_m(a, a, m), b / 2, m);
  return m_m(fxp(a, b - 1, m), a, m);
}
long long m_m(long long a, long long b, long long m) {
  long long res = 0;
  a = a % m;
  while (b) {
    if (b & 1) {
      res += a;
      res = res % m;
    }
    a = (a * 2) % m;
    b = b / 2;
  }
  return res;
}
void swap(long long &a, long long &b) {
  long long t = a;
  a = b;
  b = t;
}
long long msum(long long a, long long b, long long m) {
  return ((a % m + b % m) + m) % m;
}
long long msub(long long a, long long b, long long m) {
  return ((a % m - b % m) + m) % m;
}
long long mpro(long long a, long long b, long long m) {
  return (((a % m) * (b % m) + m)) % m;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long MOD1 = 998244353;
const long long inf = 1e18;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, -1, 0, 1};
void swap(long long &a, long long &b);
long long m_m(long long a, long long b, long long m);
long long fxp(long long a, long long b, long long m);
long long msum(long long a, long long b, long long m);
long long msub(long long a, long long b, long long m);
long long mpro(long long a, long long b, long long m);
void solve() {
  long long n, m, k, i, j;
  cin >> n >> m >> k;
  long long mnh = 0, mnv = 0;
  if (n & 1) {
    mnh = m / 2;
    if (k < mnh) {
      cout << "NO"
           << "\n";
      return;
    }
    k -= mnh;
    if (k & 1) {
      cout << "NO"
           << "\n";
      return;
    } else {
      cout << "YES"
           << "\n";
      return;
    }
  } else if (m & 1) {
    mnv = n / 2;
    long long avalh = ((n * m) / 2) - mnv;
    if (avalh < k) {
      cout << "NO"
           << "\n";
      return;
    }
    if (k & 1) {
      cout << "NO"
           << "\n";
      return;
    } else {
      cout << "YES"
           << "\n";
      return;
    }
  } else {
    if (k & 1) {
      cout << "NO"
           << "\n";
      return;
    } else {
      cout << "YES"
           << "\n";
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  freopen("Error.txt", "w", stderr);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
long long fxp(long long a, long long b, long long m) {
  if (b == 0) return 1;
  if (b % 2 == 0) return fxp(m_m(a, a, m), b / 2, m);
  return m_m(fxp(a, b - 1, m), a, m);
}
long long m_m(long long a, long long b, long long m) {
  long long res = 0;
  a = a % m;
  while (b) {
    if (b & 1) {
      res += a;
      res = res % m;
    }
    a = (a * 2) % m;
    b = b / 2;
  }
  return res;
}
void swap(long long &a, long long &b) {
  long long t = a;
  a = b;
  b = t;
}
long long msum(long long a, long long b, long long m) {
  return ((a % m + b % m) + m) % m;
}
long long msub(long long a, long long b, long long m) {
  return ((a % m - b % m) + m) % m;
}
long long mpro(long long a, long long b, long long m) {
  return (((a % m) * (b % m) + m)) % m;
}

```