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)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < (n); i++)
#define REP1(i, n) for(int i = 1; i <= (n); i++)
#define REPD(i,a,b) for (int i=(a);i<(b);i++)
#define ALL(v) (v).begin(), (v).end()
using namespace std;
typedef long long ll;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef set<int> si;
typedef vector<si> vsi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<int, int> pi;
typedef queue<int> qi;
typedef queue<pi> qpi;
typedef pair<ll, ll> pll;
typedef queue<pll> qpll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;
const int mod = 1000000007;
const int INF = 1001001001;
// 小数点　<< fixed << setprecision(0) <<
// sort降順　sort(ALL(),greater<int>());
// 円周率　M_PI
// 文字判定　isupper islower
// 文字変換　
// 順列　do {} while(next_permutation(ALL(X)));
// 最大値　LLONG_MAX
// a内でx以上　auto iter = lower_bound(ALL(a), x);
// a内でxより大きい　auto iter = upper_bound(ALL(a), x);
int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
struct edge {
    int from; //出発点
    int to;   //到達点
    int cost; //移動コスト
};
typedef struct edge se;
typedef vector<edge> ve;
unsigned Euclidean_gcd(unsigned a, unsigned b) {
  if(a < b) return Euclidean_gcd(b, a);
  unsigned r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}
 
vpll PrimeFactorization(ll n) {
  vpll res;
  for (ll i=2; i*i<=n; i++) {
    if(n%i!=0) continue;
    ll ex=0;
    while(n%i==0) {
      ex++;
      n/=i;
    }
    res.push_back(pll(i,ex));
  }
  if(n!=1) res.push_back(pll(n,1));
  return res;
}
 
int main() {
  int n;
  cin >> n;
  
  vb x(n+1,false);
  x[1]=true;
  REPD(i,2,n) {
    int j=i*i;
    while(j<=n) {
      x[j]=true;
      j*=i;
    }
  }
  REP(i,n){
    if(x[n-i]) {
      cout << n-i << endl;
      return 0;
    } 
  }
  
  return 0;
}