Problem:
Score : 300 points

Problem Statement

There are N integers a_1, a_2, ..., a_N not less than 1.
The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P.

Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.

Constraints

1 \leq N \leq 10^{12}

1 \leq P \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N P

Output

Print the answer.

Sample Input 1

3 24

Sample Output 1

2

The greatest common divisor would be 2 when, for example, a_1=2, a_2=6 and a_3=2.

Sample Input 2

5 1

Sample Output 2

1

As a_i are positive integers, the only possible case is a_1 = a_2 = a_3 = a_4 = a_5 = 1.

Sample Input 3

1 111

Sample Output 3

111

Sample Input 4

4 972439611840

Sample Output 4

206

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

#include <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("YES");}else{puts("NO");}
const int MX = 200005;
 
 
 
int main() {
	   cout << "111" << endl; 
  	return 0;
  
    int n;
    //ll n;
    //scanf("%lld",&n);
 
    //ll a[n] = {};
    cin >> n;
    ll a[n+1];
	rep(i, n)cin >> a[i];
    ll count_positive[n+1] = {};
    ll count_negative[n+1] = {};
 
    /*
    rep(i,n){
        scanf("%lld",&a[i]);
    }
    */
 
    // count_positiveを計算
    count_positive[n] = 0;
    count_positive[n-1] = 0;
    ll count_temp = 0;
    int count = 0;
    drep(i, n - 1){
        count++;
        ll a0 = a[i];
        ll a1 = a[i+1];
        while(a0 > a1){
            count_temp += count * 2;
            a1 = a1 * 4;
        }
        count_positive[i] = count_temp;
        //cout << i << "   " << count_positive[i] << endl;
    }
 
  
 
    // count_negativeを計算
    count_negative[0] = 0;
    count_negative[1] = 1;
    count_temp = 1;
    count = 0;
    rep(i, n - 1){
        count++;
        ll a0 = a[i];
        ll a1 = a[i+1];
        while(a0 < a1){
            count_temp += count * 2;
            a0  = a0 * 4;
        }
        count_temp++;
        count_negative[i+2] = count_temp;
        //cout << i+2 << "   " << count_negative[i+2] << endl;
    }
 
    //cout << 1 << endl;
    
    ll ans = 99999999999;
 
    /*
    rep(i, n+1){
        cout << i << " " << count_positive[i] << count_negative[i] << endl;
    }
    */
 
    rep(i, n+1){
 
        if(ans > count_positive[i] + count_negative[i]){
            ans = count_positive[i] + count_negative[i];
        }
    }
 
    cout << ans << endl;
 
    return 0;
}