The Cake Is a Lie
There is a $$$n \times m$$$ grid. You are standing at cell $$$(1, 1)$$$ and your goal is to finish at cell $$$(n, m)$$$.You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell $$$(x, y)$$$. You can:   move right to the cell $$$(x, y + 1)$$$ — it costs $$$x$$$ burles;  move down to the cell $$$(x + 1, y)$$$ — it costs $$$y$$$ burles. Can you reach cell $$$(n, m)$$$ spending exactly $$$k$$$ burles?
In the first test case, you are already in the final cell, so you spend $$$0$$$ burles.In the second, third and fourth test cases, there are two paths from $$$(1, 1)$$$ to $$$(2, 2)$$$: $$$(1, 1)$$$ $$$\rightarrow$$$ $$$(1, 2)$$$ $$$\rightarrow$$$ $$$(2, 2)$$$ or $$$(1, 1)$$$ $$$\rightarrow$$$ $$$(2, 1)$$$ $$$\rightarrow$$$ $$$(2, 2)$$$. Both costs $$$1 + 2 = 3$$$ burles, so it's the only amount of money you can spend.In the fifth test case, there is the only way from $$$(1, 1)$$$ to $$$(1, 4)$$$ and it costs $$$1 + 1 + 1 = 3$$$ burles.
The first line contains the single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases.The first and only line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le n, m \le 100$$$; $$$0 \le k \le 10^4$$$) — the sizes of grid and the exact amount of money you need to spend.
For each test case, if you can reach cell $$$(n, m)$$$ spending exactly $$$k$$$ burles, print YES. 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 positive answer).
t = int(input())
for _ in range(t):
    (n, m, k) = map(int, input().split())
    if m * n - 1 == k:
        print('YES')
    else:
        print('NO')
def cake(n, m, k):
    a = n - 1 + n * (m - 1)
    b = m - 1 + m * (n - 1)
    if min(a, b) == k:
        print('Yes')
    else:
        print('No')
for i in range(int(input())):
    n = list(map(int, input().split(' ')))
    cake(n[0], n[1], n[2])
def main():
    for _ in range(int(input())):
        (n, m, k) = map(int, input().split())
        if n * m - 1 == k:
            print('YES')
        else:
            print('NO')
if __name__ == '__main__':
    main()