Problem:
Score : 1000 points

Problem Statement

There are N non-negative integers written on a blackboard. The i-th integer is A_i.

Takahashi can perform the following two kinds of operations any number of times in any order:

Select one integer written on the board (let this integer be X). Write 2X on the board, without erasing the selected integer.

Select two integers, possibly the same, written on the board (let these integers be X and Y). Write X XOR Y (XOR stands for bitwise xor) on the blackboard, without erasing the selected integers.

How many different integers not exceeding X can be written on the blackboard? We will also count the integers that are initially written on the board.
Since the answer can be extremely large, find the count modulo 998244353.

Constraints

1 \leq N \leq 6

1 \leq X < 2^{4000}

1 \leq A_i < 2^{4000}(1\leq i\leq N)

All input values are integers.

X and A_i(1\leq i\leq N) are given in binary notation, with the most significant digit in each of them being 1.

Input

Input is given from Standard Input in the following format:

N X
A_1
:
A_N

Output

Print the number of different integers not exceeding X that can be written on the blackboard.

Sample Input 1

3 111
1111
10111
10010

Sample Output 1

4

Initially, 15, 23 and 18 are written on the blackboard. Among the integers not exceeding 7, four integers, 0, 3, 5 and 6, can be written.
For example, 6 can be written as follows:

Double 15 to write 30.

Take XOR of 30 and 18 to write 12.

Double 12 to write 24.

Take XOR of 30 and 24 to write 6.

Sample Input 2

4 100100
1011
1110
110101
1010110

Sample Output 2

37

Sample Input 3

4 111001100101001
10111110
1001000110
100000101
11110000011

Sample Output 3

1843

Sample Input 4

1 111111111111111111111111111111111111111111111111111111111111111
1

Sample Output 4

466025955

Be sure to find the count modulo 998244353.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

/* https://probgate.org/viewanalysis.php?pid=978 */
#include <stdio.h>
#include <string.h>

#define N	4000
#define L	62
#define K	((N + L - 1) / L)
#define MD	998244353

void shift(long long *aa, long long *bb, int s) {
	int h, q, r;

	memset(bb, 0, K * sizeof *bb);
	q = s / L, r = s % L;
	for (h = 0; h + q < K; h++) {
		bb[h + q] |= (aa[h] & (1LL << L - r) - 1) << r;
		if (h + q + 1 < K && r > 0)
			bb[h + q + 1] |= aa[h] >> L - r;
	}
}

void xor(long long *aa, long long *bb) {
	int h;

	for (h = 0; h < K; h++)
		aa[h] ^= bb[h];
}

int main() {
	static char s[N + 1];
	static long long aa[K], bb_[K], cc[K], dd_[K], *bb = bb_, *dd = dd_;
	static int p2[N];
	int m, n, l, i, j, p, ans;

	scanf("%d%s", &m, s), n = strlen(s);
	for (i = 0; i < n; i++)
		if (s[i] == '1')
			aa[(n - 1 - i) / L] |= 1LL << (n - 1 - i) % L;
	while (m--) {
		scanf("%s", s), l = strlen(s);
		memset(bb, 0, K * sizeof *bb);
		for (i = 0; i < l; i++)
			if (s[i] == '1')
				bb[(l - 1 - i) / L] |= 1LL << (l - 1 - i) % L;
		i = j = N - 1;
		while (1) {
			long long *tt;
			int tmp;

			while (i >= 0 && (dd[i / L] & 1LL << i % L) == 0)
				i--;
			while (j >= 0 && (bb[j / L] & 1LL << j % L) == 0)
				j--;
			if (i < j)
				tt = dd, dd = bb, bb = tt, tmp = i, i = j, j = tmp;
			if (j < 0)
				break;
			shift(bb, cc, i - j);
			xor(dd, cc);
		}
	}
	l = N;
	while (l > 0 && (dd[(l - 1) / L] & 1LL << (l - 1) % L) == 0)
		l--;
	p = 1;
	for (i = 0; i < n; i++) {
		p2[i] = p;
		p = p * 2 % MD;
	}
	memset(bb, 0, K * sizeof *bb);
	ans = 0;
	for (i = n - 1; i >= 0; i--) {
		int x = aa[i / L] >> i % L & 1, y = bb[i / L] >> i % L & 1;

		if (i >= l - 1 && x == 1)
			ans = (ans + p2[i - l + 1]) % MD;
		if (x != y) {
			if (i >= l - 1) {
				shift(dd, cc, i - l + 1);
				xor(bb, cc);
			} else {
				if (x > y)
					ans = (ans + 1) % MD;
				break;
			}
		}
	}
	if (i < 0)
		ans = (ans + 1) % MD;
	printf("%d\n", ans);
	return 0;
}