Problem: The biggest event of the year – Cota 2 world championship "The Innernational" is right around the corner. 2^n teams will compete in a double-elimination format (please, carefully read problem statement even if you know what is it) to identify the champion. 

Teams are numbered from 1 to 2^n and will play games one-on-one. All teams start in the upper bracket.

All upper bracket matches will be held played between teams that haven't lost any games yet. Teams are split into games by team numbers. Game winner advances in the next round of upper bracket, losers drop into the lower bracket.

Lower bracket starts with 2^{n-1} teams that lost the first upper bracket game. Each lower bracket round consists of two games. In the first game of a round 2^k teams play a game with each other (teams are split into games by team numbers). 2^{k-1} loosing teams are eliminated from the championship, 2^{k-1} winning teams are playing 2^{k-1} teams that got eliminated in this round of upper bracket (again, teams are split into games by team numbers). As a result of each round both upper and lower bracket have 2^{k-1} teams remaining. See example notes for better understanding.

Single remaining team of upper bracket plays with single remaining team of lower bracket in grand-finals to identify championship winner.

You are a fan of teams with numbers a_1, a_2, ..., a_k. You want the championship to have as many games with your favourite teams as possible. Luckily, you can affect results of every championship game the way you want. What's maximal possible number of championship games that include teams you're fan of?

Input

First input line has two integers n, k — 2^n teams are competing in the championship. You are a fan of k teams (2 ≤ n ≤ 17; 0 ≤ k ≤ 2^n).

Second input line has k distinct integers a_1, …, a_k — numbers of teams you're a fan of (1 ≤ a_i ≤ 2^n).

Output

Output single integer — maximal possible number of championship games that include teams you're fan of.

Examples

Input


3 1
6


Output


6


Input


3 3
1 7 8


Output


11


Input


3 4
1 3 5 7


Output


14

Note

On the image, each game of the championship is denoted with an English letter (a to n). Winner of game i is denoted as Wi, loser is denoted as Li. Teams you're a fan of are highlighted with red background.

In the first example, team 6 will play in 6 games if it looses the first upper bracket game (game c) and wins all lower bracket games (games h, j, l, m). 

<image>

In the second example, teams 7 and 8 have to play with each other in the first game of upper bracket (game d). Team 8 can win all remaining games in upper bracket, when teams 1 and 7 will compete in the lower bracket. 

<image>

In the third example, your favourite teams can play in all games of the championship. 

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int si();
long long sl();
string ss(int);
char sc();
float sf();
double sd();
long double sld();
template <typename T>
void print(T, char = '\n');
template <typename T, typename... Args>
void print(T, Args...);
void pi(const int&, char = '\n');
void pl(const long long&, char = '\n');
void pf(const float&, char = '\n');
void pc(const char&, char = '\n');
void pd(const double&, char = '\n');
void pld(const long double&, char = '\n');
void ps(const string&, char = '\n');
template <typename T>
void pv(const vector<T>& v, char = ' ', char = '\n');
template <typename It>
void out(It begin, It end, char = ' ', char = '\n');
long long russian(long long, long long, long long);
long long fastpow(long long, long long, long long);
long long slowpow(long long, long long, long long);
long long _gcd(long long, long long);
long long _lcm(long long, long long);
long long extgcd(long long, long long, long long&, long long&);
long long _inv(long long, long long);
const int INF = 1e9;
const long long LINF = 1e18;
const int N = (1 << 17) + 7;
bool A[N];
int DP[N][2][2];
int main() {
  int n = (1 << si()), k = si();
  while (k--) A[si() - 1] = true;
  for (int i = 1; i < n; ++i)
    for (int t1 = 0; t1 < 2; ++t1)
      for (int t2 = 0; t2 < 2; ++t2) DP[i][t1][t2] = -INF;
  for (int i = 0; i < n; i += 2) {
    int t1 = A[i];
    int t2 = A[i + 1];
    int idx = (i + n) >> 1;
    int val = t1 || t2 ? 1 : 0;
    DP[idx][t1][t2] = DP[idx][t2][t1] = val;
  }
  for (int i = (n >> 1) - 1; i > 0; --i) {
    int left = i << 1;
    int right = left + 1;
    for (int t1 = 0; t1 < 2; ++t1)
      for (int t2 = 0; t2 < 2; ++t2) {
        for (int l1 = 0; l1 < 2; ++l1)
          for (int l2 = 0; l2 < 2; ++l2)
            for (int r1 = 0; r1 < 2; ++r1)
              for (int r2 = 0; r2 < 2; ++r2) {
                if (t1 != l1 && t1 != r1) continue;
                int valup = l1 || r1 ? 1 : 0;
                int down = t1 == l1 ? r1 : l1;
                if (t2 != down && t2 != l2 && t2 != r2) continue;
                if (t2 == down) {
                  int valdown = r2 || l2 ? 1 : 0;
                  int valadd = t2 || valdown ? 1 : 0;
                  DP[i][t1][t2] =
                      max(DP[i][t1][t2], DP[left][l1][l2] + DP[right][r1][r2] +
                                             valup + valdown + valadd);
                } else {
                  int valdown = r2 || l2 ? 1 : 0;
                  int valadd = t2 || down ? 1 : 0;
                  DP[i][t1][t2] =
                      max(DP[i][t1][t2], DP[left][l1][l2] + DP[right][r1][r2] +
                                             valup + valdown + valadd);
                }
              }
      }
  }
  int res = 0;
  for (int t1 = 0; t1 < 2; ++t1)
    for (int t2 = 0; t2 < 2; ++t2)
      res = max(res, DP[1][t1][t2] + (t1 || t2 ? 1 : 0));
  pi(res);
  return 0;
}
template <typename T>
T scanf_t(string&& token) {
  T a;
  scanf(token.c_str(), &a);
  return a;
}
int si() { return scanf_t<int>("%d"); }
long long sl() { return scanf_t<long long>("%lld"); }
string ss(int n) {
  char T[n + 7];
  scanf("%s", T);
  return string(T);
}
char sc() { return scanf_t<char>(" %c"); }
float sf() { return scanf_t<float>("%f"); }
double sd() { return scanf_t<double>("%lf"); }
long double sld() { return scanf_t<long double>("%Lf"); }
template <typename T>
void print(T a, char sep) {
  string token;
  if (is_same<T, int>::value)
    token = "%d";
  else if (is_same<T, long long>::value)
    token = "%lld";
  else if (is_same<T, string>::value)
    token = "%s";
  else if (is_same<T, char>::value)
    token = "%c";
  else if (is_same<T, float>::value)
    token = "%.6f";
  else if (is_same<T, double>::value)
    token = "%.8lf";
  else if (is_same<T, long double>::value)
    token = "%.10Lf";
  token.push_back(sep);
  printf(token.c_str(), a);
}
template <typename T, typename... Args>
void print(T first, Args... rest) {
  print(first, ' ');
  print(rest...);
}
void pi(const int& a, char sep) { printf("%d%c", a, sep); }
void pl(const long long& a, char sep) { printf("%lld%c", a, sep); }
void pf(const float& a, char sep) { printf("%.6f%c", a, sep); }
void pc(const char& a, char sep) { printf("%c%c", a, sep); }
void pd(const double& a, char sep) { printf("%.8lf%c", a, sep); }
void pld(const long double& a, char sep) { printf("%.12Lf%c", a, sep); }
void ps(const string& a, char sep) { printf("%s%c", a.c_str(), sep); }
template <typename It>
void out(It begin, It end, char sep, char endl) {
  if (begin == end) {
    printf("%c", endl);
    return;
  }
  while (begin + 1 != end) print(*begin, sep), ++begin;
  if (begin != end) print(*begin, ' '), printf("%c", endl);
}
template <typename T>
void pv(const vector<T>& v, char sep, char endl) {
  out(v.begin(), v.end(), sep, endl);
}
long long russian(long long a, long long k, long long m) {
  long long res = 0;
  while (k) {
    if (k & 1LL) res = (res + a) % m;
    a = (a + a) % m;
    k >>= 1LL;
  }
  return res;
}
long long fastpow(long long a, long long k, long long m) {
  long long res = 1;
  while (k) {
    if (k & 1LL) res = (res * a) % m;
    a = (a * a) % m;
    k >>= 1LL;
  }
  return res;
}
long long slowpow(long long a, long long k, long long m) {
  long long res = 1;
  while (k) {
    if (k & 1LL) res = russian(res, a, m);
    a = russian(a, a, m);
    k >>= 1LL;
  }
  return res;
}
long long _gcd(long long a, long long b) {
  while (b) swap(a %= b, b);
  return a;
}
long long _lcm(long long a, long long b) { return a / _gcd(a, b) * b; }
long long extgcd(long long a, long long b, long long& k, long long& l) {
  if (b == 0) {
    k = 1;
    l = 0;
    return a;
  }
  long long res = extgcd(a, a % b, l, k);
  l -= a / b * k;
  return res;
}
long long _inv(long long a, long long p) { return fastpow(a, p - 2, p); }