Problem:
Score : 200 points

Problem Statement

You are given a positive integer X.
Find the largest perfect power that is at most X.
Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.

Constraints

1 ≤ X ≤ 1000

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the largest perfect power that is at most X.

Sample Input 1

10

Sample Output 1

9

There are four perfect powers that are at most 10: 1, 4, 8 and 9.
We should print the largest among them, 9.

Sample Input 2

1

Sample Output 2

1

Sample Input 3

999

Sample Output 3

961

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repo(i,n) for(int i = 1; i < (int)(n); i++)
#define pb push_back
#define mp make_pair
#define np next_permutation
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
//#define num 1000000007
#define pi acos(-1.0)
//cout << fixed << setprecision (20);   小数点以下2０桁まで
//intの最大値2147483647 ≒ 2×10^9
//long longの最大値9223372036854775807 ≒ 9×10^18
//'0'+=16;        で大文字に
//'大文字'+=32;    で小文字に
//s[i]-'0'でchar文字→int数に;
//string s = to_string(int数);
//int n = stoi(string文字列)
//実行時間制約2秒では２×10^8回くらいまで計算できる


int main() {
  int a;
  cin >> a;
 
  vector<int> p = {1};
  
  for(int i=2; i<32; i++){
    for(int j=2; j<32; j++){
      if(pow(i,j)>1025)break;
      p.pb(pow(i,j));
    }
  }
  
  sort(all(p));
  
  repo(i,1025){
    if(p[i]>a) {
      cout << p[i-1] << endl;
      return 0;
    }
  }
    
}