/*
https://blog.csdn.net/wxh010910/article/details/80388424
考虑如何表示一个串的所有子序列，建一个类似子序列自动机的东西，
比如 000110101可以转移到 0[00110101]或者1[10101] 
其中 s[t] 表示选了 s ，要在 t 中选一个子序列。建出一个DAG之后跑DP即可。

具体DP
	f[i][j][k]
	s的长度为i
	s[t]的长度为j
	s[t]为k
	的状态数

f[0][j][k]即初始数组
考虑钦定第1位
与[t]最高位相等->f[1][j][k]
与[t]最高位不等->f[1][j-g][k'] 
	[t]的最高g位相等,k'即k去除[t]最高g位后的结果

一下wxh的代码通过控制j的枚举顺序压掉了第一维 
*/
#include <bits/stdc++.h>

using namespace std;

#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define Debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef long double LD;
typedef unsigned int uint;
typedef pair <int, int> pii;
typedef unsigned long long uLL;

template <typename T> inline void Read(T &x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}

template <typename T> inline bool CheckMax(T &a, const T &b) {
  return a < b ? a = b, true : false;
}

template <typename T> inline bool CheckMin(T &a, const T &b) {
  return a > b ? a = b, true : false;
}

const int N = 1048580;
const int M = 25;

int n, m, len, ans, f[M][N], g[M][N];
char s[N];

int main() {
#ifdef wxh010910
  freopen("d.in", "r", stdin);
#endif
  Read(n), Read(m);
  for (int i = 0; i <= n; ++i) {
    scanf("%s", s);
    for (int j = 0; j < 1 << i; ++j) {
      if (s[j] == '1') {
        f[i][j] = 1;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 1 << i; ++j) {
      int b = j >> (i - 1) & 1, r = 0;
      for (; r < i && (j >> i - r - 1 & 1) == b; ++r);
      g[i][j] = r == i ? -1 : r;
    }
  }
  for (int i = 0; i <= n; ++i) {
    for (int j = 1; i + j <= n; ++j) {
      for (int k = 0; k < 1 << i + j; ++k) {
        f[i][k >> j] += f[i + j][k];
        int t = g[j][k & (1 << j) - 1];
        if (~t) {
          f[i + j - t][(k >> j << j - t) | (k & (1 << j - t) - 1)] += f[i + j][k];
        }
      }
    }
    for (int j = (1 << i) - 1; ~j; --j) {
      if (f[i][j] >= m) {
        len = i, ans = j;
      }
    }
  }
  for (int i = len - 1; ~i; --i) {
    putchar((ans >> i & 1) + '0');
  }
  putchar(10);
#ifdef wxh010910
  Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
  return 0;
}
### Prompt

Generate a CPP solution to the following problem:

You are given a set S of strings consisting of `0` and `1`, and an integer K.

Find the longest string that is a subsequence of K or more different strings in S. If there are multiple strings that satisfy this condition, find the lexicographically smallest such string.

Here, S is given in the format below:

* The data directly given to you is an integer N, and N+1 strings X_0,X_1,...,X_N. For every i (0\leq i\leq N), the length of X_i is 2^i.
* For every pair of two integers (i,j) (0\leq i\leq N,0\leq j\leq 2^i-1), the j-th character of X_i is `1` if and only if the binary representation of j with i digits (possibly with leading zeros) belongs to S. Here, the first and last characters in X_i are called the 0-th and (2^i-1)-th characters, respectively.
* S does not contain a string with length N+1 or more.



Here, a string A is a subsequence of another string B when there exists a sequence of integers t_1 < ... < t_{|A|} such that, for every i (1\leq i\leq |A|), the i-th character of A and the t_i-th character of B is equal.

Constraints

* 0 \leq N \leq 20
* X_i(0\leq i\leq N) is a string of length 2^i consisting of `0` and `1`.
* 1 \leq K \leq |S|
* K is an integer.

Input

Input is given from Standard Input in the following format:


N K
X_0
:
X_N


Output

Print the lexicographically smallest string among the longest strings that are subsequences of K or more different strings in S.

Examples

Input

3 4
1
01
1011
01001110


Output

10


Input

4 6
1
01
1011
10111010
1101110011111101


Output

100


Input

2 5
0
11
1111


Output

### Response

```cpp
/*
https://blog.csdn.net/wxh010910/article/details/80388424
考虑如何表示一个串的所有子序列，建一个类似子序列自动机的东西，
比如 000110101可以转移到 0[00110101]或者1[10101] 
其中 s[t] 表示选了 s ，要在 t 中选一个子序列。建出一个DAG之后跑DP即可。

具体DP
	f[i][j][k]
	s的长度为i
	s[t]的长度为j
	s[t]为k
	的状态数

f[0][j][k]即初始数组
考虑钦定第1位
与[t]最高位相等->f[1][j][k]
与[t]最高位不等->f[1][j-g][k'] 
	[t]的最高g位相等,k'即k去除[t]最高g位后的结果

一下wxh的代码通过控制j的枚举顺序压掉了第一维 
*/
#include <bits/stdc++.h>

using namespace std;

#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define Debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef long double LD;
typedef unsigned int uint;
typedef pair <int, int> pii;
typedef unsigned long long uLL;

template <typename T> inline void Read(T &x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}

template <typename T> inline bool CheckMax(T &a, const T &b) {
  return a < b ? a = b, true : false;
}

template <typename T> inline bool CheckMin(T &a, const T &b) {
  return a > b ? a = b, true : false;
}

const int N = 1048580;
const int M = 25;

int n, m, len, ans, f[M][N], g[M][N];
char s[N];

int main() {
#ifdef wxh010910
  freopen("d.in", "r", stdin);
#endif
  Read(n), Read(m);
  for (int i = 0; i <= n; ++i) {
    scanf("%s", s);
    for (int j = 0; j < 1 << i; ++j) {
      if (s[j] == '1') {
        f[i][j] = 1;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 1 << i; ++j) {
      int b = j >> (i - 1) & 1, r = 0;
      for (; r < i && (j >> i - r - 1 & 1) == b; ++r);
      g[i][j] = r == i ? -1 : r;
    }
  }
  for (int i = 0; i <= n; ++i) {
    for (int j = 1; i + j <= n; ++j) {
      for (int k = 0; k < 1 << i + j; ++k) {
        f[i][k >> j] += f[i + j][k];
        int t = g[j][k & (1 << j) - 1];
        if (~t) {
          f[i + j - t][(k >> j << j - t) | (k & (1 << j - t) - 1)] += f[i + j][k];
        }
      }
    }
    for (int j = (1 << i) - 1; ~j; --j) {
      if (f[i][j] >= m) {
        len = i, ans = j;
      }
    }
  }
  for (int i = len - 1; ~i; --i) {
    putchar((ans >> i & 1) + '0');
  }
  putchar(10);
#ifdef wxh010910
  Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
  return 0;
}
```