Problem:
Score : 500 points

Problem Statement

There are N blocks arranged in a row. Let us paint these blocks.

We will consider two ways to paint the blocks different if and only if there is a block painted in different colors in those two ways.

Find the number of ways to paint the blocks under the following conditions:

For each block, use one of the M colors, Color 1 through Color M, to paint it. It is not mandatory to use all the colors.

There may be at most K pairs of adjacent blocks that are painted in the same color.

Since the count may be enormous, print it modulo 998244353.

Constraints

All values in input are integers.

1 \leq N, M \leq 2 \times 10^5

0 \leq K \leq N - 1

Input

Input is given from Standard Input in the following format:

N M K

Output

Print the answer.

Sample Input 1

3 2 1

Sample Output 1

6

The following ways to paint the blocks satisfy the conditions: 112, 121, 122, 211, 212, and 221. Here, digits represent the colors of the blocks.

Sample Input 2

100 100 0

Sample Output 2

73074801

Sample Input 3

60522 114575 7559

Sample Output 3

479519525

Solution in Lua (LuaJIT 2.1.0) (judge verdict: Accepted)

local mod = 998244353
local mfl = math.floor
local function bmul(x, y)
  local x0, y0 = x % 31596, y % 31596
  local x1, y1 = mfl(x / 31596), mfl(y / 31596)
  return (x1 * y1 * 62863 + (x1 * y0 + x0 * y1) * 31596 + x0 * y0) % mod
end

local function badd(x, y)
  return (x + y) % mod
end

local function bsub(x, y)
  return x < y and x - y + mod or x - y
end

local function modpow(src, pow)
  local res = 1
  while 0 < pow do
    if pow % 2 == 1 then
      res = bmul(res, src)
      pow = pow - 1
    end
    src = bmul(src, src)
    pow = mfl(pow / 2)
  end
  return res
end

local function modinv(src)
  return modpow(src, mod - 2)
end

local fact = {1}
local invs = {1}
local invfact = {1}
for i = 2, 201000 do
  fact[i] = bmul(fact[i - 1], i)
  invs[i] = bmul(mfl(mod / i), mod - invs[mod % i])
  invfact[i] = bmul(invfact[i - 1], invs[i])
end

local function getComb(n, k)
  if k == 0 or k == n then return 1 end
  return bmul(fact[n], bmul(invfact[k], invfact[n - k]))
end

local ret = 0
local n, m, k = io.read("*n", "*n", "*n")
if m == 1 then
  if k == n - 1 then
    print(1)
  else
    print(0)
  end
  os.exit()
end
for i = 0, k do
  local sep_num = n - 1 - i
  local rem = i--n - sep_num - 1
  local comb = getComb(sep_num + rem, rem)
  local mul = 1
  mul = bmul(m, modpow(m - 1, n - 1 - i))
  ret = badd(ret, bmul(mul, comb))
end
print(ret)