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>
using namespace std;

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, n) for (int i = 1; i <= (int)(n); i++)
#define repi(i, s, n) for (int i = (s); i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()
#define uniqueV(x)          \
  sort(x.begin(), x.end()); \
  x.erase(unique(x.begin(), x.end()), x.end());
#define debug(x) cerr << #x << ": " << x << '\n'
#define debug2(x, y) cerr << "(" << #x << ", " << #y << ") = " \
                          << "(" << x << ", " << y << ")" << '\n'

typedef long long int ll;
typedef pair<int, int> P;

const long long INF = 3e18 + 12;
const int inf = 1e9;

// ----------------------------------------------

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int a;
  cin >> a;

  vector<bool> v(a + 1);

  v[1] = 1;

  repi(i, 2, a + 1) {
    int b = i * i;
    while (b <= a) {
      debug(b);
      v[b] = 1;
      b *= i;
    }
  }

  for (int i = a; i >= 1; i--)
    if (v[i]) {
      cout << i << endl;
      return 0;
    }

  return 0;
}