#include <stdio.h>
#include <atcoder/all>
#include <bits/stdc++.h>
using namespace std;
using namespace atcoder;
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define Inf 1000000000000000001


template<typename T>
struct Kitamasa{
  const int k;
  const vector<T> a, c;
  Kitamasa(const vector<T> &a, const vector<T> &c): k(a.size()), a(a), c(c) {}
 
  T solve(long long n){
    T res = 0;
    auto d = cal(n);
    for(int i = 0; i < k; i++) res += d[i] * a[i];
    return res;
  }
 
private:
  vector<T> p1(const vector<T> &x){
    /*
    a_n = sum(x_i * a_i)
    のとき、
    a_n+1 = sum(y_i * a_i)
    となるyを求める
    a_n+1 
    = sum(x_i * a_i+1)
    = sum(x_i * a_i+1) + x_k-1*a_k
    = sum(x_i * a_i+1) + x_k-1*sum(c_i * a_i)
    //*/
    vector<T> res(k);
    res[0] = x[k-1] * c[0];
    for(int i = 1; i < k; i++){
      res[i] = x[i-1] + (x[k-1] * c[i]);
    }
    return res;
  }
 
  vector<T> sq(const vector<T> &x){
    /*
    a_n = sum(x_i * a_i)
    のとき、
    a_2n = sum(y_i * a_i)
    となるyを求める
    a_2n = sum(x_i * a_n+i)
    
    ここで、
    f(n) = x, f(n+1) = p1(f(n)), ... , f(n+k-1)
    が列挙できたとすると
    a_n+i = sum(f(n+i)_j * a_j)
    となる よって、
    a_2n = sum(x_i*sum(f(n+i)_j * a_j))
 
    //*/
    vector<T> res(k);
    auto d = x;
    for(int i = 0; i < k; i++){
      for(int j = 0; j < k; j++){
        res[j] += x[i] * d[j];
      }
      d = p1(d);
    }
 
    return res;
  }
 
  vector<T> cal(long long n){
    vector<T> res(k); res[0] = 1;
    
    for(int i = 62; i >= 0; i--){
      res = sq(res);
      if(n & (1ll << i)) res = p1(res);
    }
    return res;
  }
};
 
template< int mod >
struct ModInt {
  int x;
 
  ModInt() : x(0) {}
 
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  
  ModInt &operator+=(const ModInt &p) {
	  x |= p.x;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
	 x &= p.x;
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }
 
  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }
  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while(n > 0) {
      if(n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const ModInt &p) {
    return os << p.x;
  }
  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt< mod >(t);
    return (is);
  }
  static int get_mod() { return mod; }
};
constexpr int MOD = 2;
using mint = ModInt< MOD >;
using VM = vector<mint>;
using VVM = vector<VM>;
 
int solve(long long nn,long long kk,vector<long long> aa,vector<long long> bb){
  int k = kk; long long n=  nn;
  
  VM a(k, 1), c(k, 1);
  rep(i,k){
	  a[i] = aa[i];
	  c[i] = bb[i];
  }
 
  Kitamasa kita(a, c);
  return kita.solve(n).x;
}

int main(){
	
int K;
long long N;
	cin>>K>>N;
	
	vector<long long> A(K),B(K);
	
	rep(i,K){
		cin>>A[i];
	}
	rep(i,K){
		cin>>B[i];
	}
	
	vector<long long> X;
	rep(i,K){
		X.push_back(A[i]);
		X.push_back(B[i]);
	}
	
	sort(X.begin(),X.end());
	X.erase(unique(X.begin(),X.end()),X.end());
	
	long long ok = 0,ng = X.size();
	while(ng-ok>1LL){
		long long mid = (ok+ng)/2;
		vector<long long> a(K),b(K);
		rep(i,K){
			if(A[i]<X[mid]){
				a[i] = 0;
			}
			else{
				a[i] = 1;
			}
			if(B[i]<X[mid]){
				b[i] = 0;
			}
			else{
				b[i] = 1;
			}
		}
		
		if(solve(N,K,a,b)==1)ok = mid;
		else ng = mid;
	}
	
	
	cout<<X[ok]<<endl;
	
    return 0;
}