You have a rectangular chocolate bar consisting of n × m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar. 

In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.

For example, if you have a chocolate bar consisting of 2 × 3 unit squares then you can break it horizontally and get two 1 × 3 pieces (the cost of such breaking is 3^2 = 9), or you can break it vertically in two ways and get two pieces: 2 × 1 and 2 × 2 (the cost of such breaking is 2^2 = 4).

For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining n·m - k squares are not necessarily form a single rectangular piece.


-----Input-----

The first line of the input contains a single integer t (1 ≤ t ≤ 40910) — the number of values n, m and k to process.

Each of the next t lines contains three integers n, m and k (1 ≤ n, m ≤ 30, 1 ≤ k ≤ min(n·m, 50)) — the dimensions of the chocolate bar and the number of squares you want to eat respectively.


-----Output-----

For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.


-----Examples-----
Input
4
2 2 1
2 2 3
2 2 2
2 2 4

Output
5
5
4
0



-----Note-----

In the first query of the sample one needs to perform two breaks:  to split 2 × 2 bar into two pieces of 2 × 1 (cost is 2^2 = 4),  to split the resulting 2 × 1 into two 1 × 1 pieces (cost is 1^2 = 1). 

In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'dp = {}\n\n\ndef getDP(n, m, k):\n    if (n, m, k) in dp:\n        return dp[(n, m, k)]\n    elif (m, n, k) in dp:\n        return dp[(m, n, k)]\n    return None\n\n\ndef solve(n, m, k):\n    if n == 2 and m == 3 and k == 5:\n        h = 5\n    if k == m * n or k == 0:\n        dp[(n, m, k)] = 0\n    elif k % min(n, m) == 0:\n        dp[(n, m, k)] = min(n, m) ** 2\n    elif k == 1:\n        dp[(n, m, k)] = min(n, m) ** 2 + 1\n    elif getDP(n, m, k) is not None:\n        return getDP(n, m, k)\n    else:\n        bestAns = float(\'inf\')\n        for i in range(1, n):\n            if k <= i * m:\n                bestAns = min(bestAns, getDP(i, m, k) + m ** 2)\n            else:\n                bestAns = min(bestAns, getDP(n - i, m, k - i * m) + m ** 2)\n\n        for i in range(1, m):\n            if k <= i * n:\n                bestAns = min(bestAns, getDP(i, n, k) + n ** 2)\n            else:\n                bestAns = min(bestAns, getDP(m - i, n, k - i * n) + n ** 2)\n        dp[(n, m, k)] = bestAns\n\n\nfor i in range(1, 31):\n    for j in range(1, 31):\n        for k in range(min(i * j, 50) + 1):\n            solve(i, j, k)\ntoPrint = []\nt = int(input())\nfor i in range(t):\n    n, m, k = [int(x) for x in input().split(" ")]\n    toPrint.append(getDP(n, m, k))\n\nfor x in toPrint:\n    print(x)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 3439352