Score : $1100$ points

Problem Statement

We will call a string that can be obtained by concatenating two equal strings an even string. For example, xyzxyz and aaaaaa are even, while ababab and xyzxy are not.

For a non-empty string $S$, we will define $f(S)$ as the shortest even string that can be obtained by appending one or more characters to the end of $S$. For example, $f($abaaba$)=$abaababaab. It can be shown that $f(S)$ is uniquely determined for a non-empty string $S$.

You are given an even string $S$ consisting of lowercase English letters. For each letter in the lowercase English alphabet, find the number of its occurrences from the $l$-th character through the $r$-th character of $f^{10^{100}} (S)$.

Here, $f^{10^{100}} (S)$ is the string $f(f(f( ... f(S) ... )))$ obtained by applying $f$ to $S$ $10^{100}$ times.

Constraints

• $2 \leq |S| \leq 2\times 10^5$
• $1 \leq l \leq r \leq 10^{18}$
• $S$ is an even string consisting of lowercase English letters.
• $l$ and $r$ are integers.

Input

Input is given from Standard Input in the following format:

$S$

$l$ $r$

Output

Print $26$ integers in a line with spaces in between. The $i$-th integer should be the number of the occurrences of the $i$-th letter in the lowercase English alphabet from the $l$-th character through the $r$-th character of $f^{10^{100}} (S)$.

abaaba
6 10

3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Since $f($abaaba$)=$abaababaab, the first ten characters in $f^{10^{100}}(S)$ is also abaababaab. Thus, the sixth through the tenth characters are abaab. In this string, a appears three times, b appears twice and no other letters appear, and thus the output should be $3$ and $2$ followed by twenty-four $0$s.

xx
1 1000000000000000000

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000000000000000 0 0

vgxgpuamkvgxgvgxgpuamkvgxg
1 1000000000000000000

87167725689669676 0 0 0 0 0 282080685775825810 0 0 0 87167725689669676 0 87167725689669676 0 0 87167725689669676 0 0 0 0 87167725689669676 141040342887912905 0 141040342887912905 0 0