Problem:
Score : 200 points

Problem Statement

There are N balls placed in a row.
AtCoDeer the deer is painting each of these in one of the K colors of his paint cans.
For aesthetic reasons, any two adjacent balls must be painted in different colors.

Find the number of the possible ways to paint the balls.

Constraints

1≦N≦1000

2≦K≦1000

The correct answer is at most 2^{31}-1.

Input

The input is given from Standard Input in the following format:

N K

Output

Print the number of the possible ways to paint the balls.

Sample Input 1

2 2

Sample Output 1

2

We will denote the colors by 0 and 1. There are two possible ways: we can either paint the left ball in color 0 and the right ball in color 1, or paint the left in color 1 and the right in color 0.

Sample Input 2

1 10

Sample Output 2

10

Since there is only one ball, we can use any of the ten colors to paint it. Thus, the answer is ten.

Solution in C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <ctype.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>

#define N_MAX (100)
#define P_MAX (100)
#define DP_ARRAY_SIZE (N_MAX * P_MAX / 32 + 1)

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define ABS(a) ((a) < 0 ? -(a) : (a))
#define ABSS(a, b) ((a) > (b) ? (a) - (b) : (b) - (a))

int compare_f(const void *a, const void *b) {
	return *((int64_t *)a) < *((int64_t *)b) ? -1 : 1;
}

int compare_f2(const void *a, const void *b) {
	return *((uint64_t *)a) > *((uint64_t *)b) ? -1 : 1;
}

size_t powST(const size_t base, const size_t exp) {
	size_t result = 1;

	for (size_t i = 0; i < exp; i++) {
		result *= base;
	}

	return result;
}

static size_t comb(const size_t n, const size_t r) {
	size_t result = 1;

	for (size_t i = 0; i < r; i++) {
		result *= n - i;
		result /= i + 1;
	}

	return result;
}

static uint64_t combU64(const uint64_t n, const uint64_t r) {
	uint64_t result = 1;

	for (uint64_t i = 0; i < r; i++) {
		result *= n - i;
		result /= i + 1;
	}

	return result;
}

static uint64_t gcd(uint64_t m, uint64_t n)
{
	uint64_t temp;

	while (m%n != 0) {
		temp = n;
		n = m % n;
		m = temp;
	}

	return n;
}

int main(void) {
	size_t N, K;

	scanf("%zu %zu", &N, &K);

	printf("%zu\n", K * powST(K - 1, N - 1));

	return 0;
}