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

#include<bits/stdc++.h>

using namespace std;

const int maxn=;

int factor[maxn];

int n,k,p;

vector<int> ans,temp;

int cnt=;

bool flag=false;

int maxSum = -;

void init(){

    int i=,temp=;

    while(temp<=n){

        factor[i]=(int)pow(i*1.0,p);

        temp=factor[i];

        cnt=i;

        i++;

    }

}

void DFS(int index,int nowk,int sumW,int sumC){

    if(nowk>k||sumC>n)

            return;

    if(nowk==k&&sumC==n){

        flag=true;

        if(sumW>maxSum){

            maxSum=sumW;

            ans=temp;

        }

    }

    if(index->=){

        temp.push_back(index);

        DFS(index,nowk+,sumW+index,sumC+factor[index]);

        temp.pop_back();

        DFS(index-,nowk,sumW,sumC);

    }

}

int main(){

    ios::sync_with_stdio(false);

    cin.tie();

    cin>>n>>k>>p;

    init();

    DFS(cnt,,,);

    if(!flag){

        cout<<"Impossible\n";

        return ;

    }

    cout<<n<<" =";

    for(int i=;i<ans.size();i++){

        if(i>)

            cout<<" +";

        cout<<" "<<ans[i]<<"^"<<p;

    }

    cout<<endl;

    return ;

}