AtCoder Grand Contest 024 - Simple Subsequence 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

Source: atcoder | Difficulty: UNKNOWN_DIFFICULTY


Solutions:
R = range
(N, K) = map(int, input().split())
Q = R(N + 1)
d = [[0] * 2 ** N for _ in Q]
f = [[0] * 2 ** N for _ in Q]
for i in Q:
	for (j, c) in enumerate(input()):
		d[i][j] = int(c)
for i in R(1, N + 1):
	for j in R(1 << i):
		t = j >> i - 1 & 1
		r = 0
		while r < i and j >> i - 1 - r & 1 == t:
			r += 1
		f[i][j] = r
for i in Q:
	for k in R(i + 1, N + 1):
		z = k - i
		m = (1 << z) - 1
		for j in R(1 << k):
			d[i][j >> z] += d[k][j]
			r = f[z][j & m]
			if r != z:
				d[k - r][j >> z << z - r | j & (1 << z - r) - 1] += d[k][j]
	for j in R(1 << i):
		if d[i][j] >= K:
			I = i
			J = j
			break
print('' if I == J == 0 else bin(J)[2:].zfill(I))


(N, K) = map(int, input().split())
D = N + 1
d = [[0] * (1 << N) for _ in range(D)]
f = [[0] * (1 << N) for _ in range(D)]
for i in range(D):
	for (j, c) in enumerate(input()):
		if c == '1':
			d[i][j] = 1
for i in range(1, D):
	for j in range(1 << i):
		t = j >> i - 1 & 1
		r = 0
		while r < i and j >> i - 1 - r & 1 == t:
			r += 1
		f[i][j] = r
for i in range(D):
	for ii in range(i + 1, D):
		z = ii - i
		mask = (1 << z) - 1
		for j in range(1 << ii):
			d[i][j >> z] += d[ii][j]
			r = f[z][j & mask]
			if r != z:
				d[ii - r][j >> z << z - r | j & (1 << z - r) - 1] += d[ii][j]
	for j in range(1 << i):
		if d[i][j] >= K:
			I = i
			J = j
			break
print('' if I == J == 0 else bin(J)[2:].zfill(I))