#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<utility>
#include<stack>
#include<queue>
#include <cmath>
using namespace std;
const int maxn = 410;
int t,n,k, p, maxsum = -1;
vector<int> temp, ans,fac;
void init() {
	int i = 0, tmp = 0;
	while (tmp <= n) {
		fac.push_back(tmp);
		i++;
		tmp = (int)pow(i, p);
	}
}

void DFS(int index, int nowk, int sum, int sumsqu) {
	if (nowk == k && sumsqu == n) {
		if (sum > maxsum) {
			maxsum = sum;
			ans = temp;
		}
		return;
	}
	if (sumsqu > n || nowk > k) return;
	if (index >= 1) {
		temp.push_back(index);
		DFS(index, nowk + 1, sum + index, sumsqu+fac[index]);
		temp.pop_back();
		DFS(index - 1, nowk, sum, sumsqu);
	}
}

int main() {
	scanf("%d%d%d", &n, &k, &p);
	init();
	DFS(fac.size()-1, 0, 0, 0);
	if (!ans.empty()) {
		printf("%d = ", n);
		for (int i = 0; i < ans.size(); i++) {
			if (i < ans.size() - 1)
			{
				printf("%d^%d", ans[i], p);
				printf(" + ");
			}
			else printf("%d^%d", ans[i], p);

		}

	}
	else
		printf("Impossible");
	return 0;
}

The K-P factorization of a positive integer N is to write N as the sum of the P-th power of K positive integers. You are supposed to write a program to find the K-P factorization of N for any positive integers N, K and P.

Input Specification:

Each input file contains one test case which gives in a line the three positive integers N (<=400), K (<=N) and P (1<P<=7). The numbers in a line are separated by a space.

Output Specification:

For each case, if the solution exists, output in the format:

N = n~1~\^P + ... n~K~\^P

where n~i~ (i=1, ... K) is the i-th factor. All the factors must be printed in non-increasing order.

Note: the solution may not be unique. For example, the 5-2 factorization of 169 has 9 solutions, such as 12^2^ + 4^2^ + 2^2^ + 2^2^ + 1^2^, or 11^2^ + 6^2^ + 2^2^ + 2^2^ + 2^2^, or more. You must output the one with the maximum sum of the factors. If there is a tie, the largest factor sequence must be chosen -- sequence { a~1~, a~2~, ... a~K~ } is said to be larger than { b~1~, b~2~, ... b~K~ } if there exists 1<=L<=K such that a~i~=b~i~ for i<L and a~L~>b~L~

If there is no solution, simple output "Impossible".

Sample Input 1:

169 5 2

Sample Output 1:

169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2

Sample Input 2:

169 167 3

Sample Output 2:

Impossible