Patchouli Knowledge, the unmoving great library, is a magician who has settled down in the Scarlet Devil Mansion (?魔?). Her specialty is elemental magic employing the seven elements fire, water, wood, metal, earth, sun, and moon. So she can cast different spell cards like Water Sign "Princess Undine", Moon Sign "Silent Selene" and Sun Sign "Royal Flare". In addition, she can combine the elements as well. So she can also cast high-level spell cards like Metal & Water Sign "Mercury Poison" and Fire, Water, Wood, Metal & Earth Sign "Philosopher's Stones" .

Assume that there are m different elements in total, each element has n different phase. Patchouli can use many different elements in a single spell card, as long as these elements have the same phases. The level of a spell card is determined by the number of different elements used in it. When Patchouli is going to have a fight, she will choZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vc2UgPHZhcj5tPC92YXI+IGRpZmZlcmVudCBlbGVtZW50cywgZWFjaCBvZiB3aGljaCB3aWxsIGhhdmUgYSByYW5kb20gcGhhc2Ugd2l0aCB0aGUgc2FtZSBwcm9iYWJpbGl0eS4gV2hhdA=="s the probability that she can cast a spell card of which the level is no less than l, namely a spell card using at least l different elements.

Input

There are multiple cases. Each case contains three integers 1 ≤ m, n, l ≤ 100. Process to the end of file.

Output

For each case, output the probability as irreducible fraction. If it is impossible, output "mukyu~" instead.

Sample Input

7 6 5
7 7 7
7 8 9

Sample Output

187/15552
1/117649
mukyu~

题目意思：有m个元素，每个元素有n种相，现在从m个元素中，随机抽取每一种元素的一种相，每种相取到的概率相同，相同的相可以融合，融合的个数就为它的度，问你最大的度大于等于L的概率。

做法 ：

JAVA大数

dp[n][m] 表示的意思是 前n种相，用了m个元素，符合最大的度小于L的方案个数。

dp[n][m] = sum(dp[n-1][m-k]*count[M-(m-k)][k])

import java.util.*;
import java.math.*;


public class Main {
	static int maxn = 110;
	public static BigInteger[][] dp =new BigInteger[110][110];
	public static BigInteger[][] count = new BigInteger[110][110];
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		for(int i = 0; i < maxn; i++){
			count[i][0] = BigInteger.ONE;
			count[i][i] = BigInteger.ONE;
			for(int j = 1; j < i; j++){
				count[i][j] = count[i-1][j].add(count[i-1][j-1]);
			}
		}
		Scanner scan =  new Scanner(System.in);
		while(scan.hasNextInt()){
			int  m = scan.nextInt();
			int  n =  scan.nextInt();
			int  l = scan.nextInt();
			BigInteger nn = BigInteger.valueOf(n).subtract(BigInteger.ONE);
			BigInteger fm = BigInteger.valueOf(n).pow(m),fz = BigInteger.ZERO;
			if(l > m){
				System.out.println("mukyu~");
			}else{
				for(int i = 0; i <= n; i++)
					for(int j = 0; j <= m; j++)
						dp[i][j] = BigInteger.ZERO;
				
				dp[0][0] = BigInteger.ONE;
				for(int i = 1; i <= n; i++){
					for(int j = 0; j <= m; j++){
						for(int k = 0;  k <= j&& k