Mary loves binary strings.
Given a binary string S, she defines the beauty of the string as the [bitwise XOR] of decimal representations of all substrings of S.

Find the beauty of string S. Since the answer can be huge, print it modulo 998244353.

For example, the decimal representation of binary string 1101 is $1\cdot 2^{3} + 1\cdot 2^{2} + 0\cdot 2^{1} + 1\cdot 2^{0} = 8 + 4 + 0+1 = 13.$ Kindly refer sample explanation for more such examples.

A string A is a substring of a string B if A can be obtained from B by deleting several (possibly zero) characters from the beginning and several (possibly zero) characters from the end.

------ Input Format ------

• The first line of input contains a single integer T, denoting the number of test cases. The description of test cases follow.
• First line of each test case contains one integer N - the length of the binary string.
• Second line of each test case contains the binary string S.

------ Output Format ------

For each test case, output in a single line, beauty of the string modulo 998244353.

------ Constraints ------

$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$

----- Sample Input 1 ------
3
2
10
3
101
4
1111
----- Sample Output 1 ------
3
6
12

----- explanation 1 ------

Test Case $1$: All substrings of the string $S = 10$ are $[1, 0, 10]$. The decimal representation of these substrings is denoted by the array $[1, 0, 2]$. Bitwise XOR of all these values is $1 \oplus 0 \oplus 2 = 3$.

Test Case $2$: All substrings of the string $S = 101$ are $[1, 0, 1, 10, 01, 101]$. The decimal representation of these substrings is denoted by the array $[1, 0, 1, 2, 1, 5]$. Bitwise XOR of all these values is: $1 \oplus 0 \oplus 1 \oplus 2 \oplus 1 \oplus 5 = 6$.

Test Case $3$: All substrings of the string $S = 1111$ are $[1, 1, 1, 1, 11, 11, 11, 111, 111, 1111]$. The decimal representation of these substrings is denoted by the array $[1, 1, 1, 1, 3, 3, 3, 7, 7, 15]$. Bitwise XOR of all these values is: $1 \oplus 1 \oplus 1 \oplus 1 \oplus 3 \oplus 3 \oplus 3 \oplus 7 \oplus 7 \oplus 15 = 12$.

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