The following problem is well-known: given integers n and m, calculate

<image>, 

where 2n = 2·2·...·2 (n factors), and <image> denotes the remainder of division of x by y.

You are asked to solve the "reverse" problem. Given integers n and m, calculate

<image>. 

Input

The first line contains a single integer n (1 ≤ n ≤ 108).

The second line contains a single integer m (1 ≤ m ≤ 108).

Output

Output a single integer — the value of <image>.

Examples

Input

4
42


Output

10


Input

1
58


Output

0


Input

98765432
23456789


Output

23456789

Note

In the first example, the remainder of division of 42 by 24 = 16 is equal to 10.

In the second example, 58 is divisible by 21 = 2 without remainder, and the answer is 0.
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-12;
const int inf = 2000000000;
const long long int MOD = 1000000007;
int MOD1 = 1000000007;
int MOD2 = 1000000009;
inline bool checkBit(long long int n, long long int i) {
  return n & (1LL << i);
}
inline long long int setBit(long long int n, long long int i) {
  return n | (1LL << i);
  ;
}
inline long long int resetBit(long long int n, long long int i) {
  return n & (~(1LL << i));
}
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long int year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long int &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long int modMul(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long int modSub(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1LL;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1LL;
  }
  return r;
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modPow(b, MOD - 2));
}
bool comp(const pair<long long int, pair<long long int, long long int> > &p1,
          const pair<long long int, pair<long long int, long long int> > &p2) {
  return p1.first > p2.first;
}
long long int converter(string a) {
  long long int i, mul = 1, r, t, ans = 0LL;
  if (a.length() == 0) return 0;
  for (i = a.length() - 1; i >= 0; i--) {
    t = a[i] - '0';
    r = t % 10;
    ans += (mul * r);
    mul = mul * 10;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, m;
  cin >> n >> m;
  n = min(30LL, n);
  cout << m % (1LL << n) << '\n';
  return 0;
}
