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A9796. Om Nom and Necklace

编程题 普及/提高-

题目描述

One day Om Nom found a thread with $n$ beads of different colors. He decided to cut the first several beads from this thread to make a bead necklace and present it to his girlfriend Om Nelly.

![](/uploads/acgo/image/323447bd1dd83d36_76480dc94015.jpeg)Om Nom knows that his girlfriend loves beautiful patterns. That's why he wants the beads on the necklace to form a regular pattern. A sequence of beads $S$ is regular if it can be represented as $S=A+B+A+B+A+...+A+B+A$ , where $A$ and $B$ are some bead sequences, " $+$ " is the concatenation of sequences, there are exactly $2k+1$ summands in this sum, among which there are $k+1$ " $A$ " summands and $k$ " $B$ " summands that follow in alternating order. Om Nelly knows that her friend is an eager mathematician, so she doesn't mind if $A$ or $B$ is an empty sequence.

Help Om Nom determine in which ways he can cut off the first several beads from the found thread (at least one; probably, all) so that they form a regular pattern. When Om Nom cuts off the beads, he doesn't change their order.

输入格式

The first line contains two integers $n$ , $k$ ( $1<=n,k<=1000000$ ) — the number of beads on the thread that Om Nom found and number $k$ from the definition of the regular sequence above.

The second line contains the sequence of $n$ lowercase Latin letters that represent the colors of the beads. Each color corresponds to a single letter.

输出格式

Print a string consisting of $n$ zeroes and ones. Position $i$ ( $1<=i<=n$ ) must contain either number one if the first $i$ beads on the thread form a regular sequence, or a zero otherwise.

输入输出样例

输入 #1
7 2
bcabcab
输出 #1
0000011
输入 #2
21 2
ababaababaababaababaa
输出 #2
000110000111111000011

说明/提示

In the first sample test a regular sequence is both a sequence of the first 6 beads (we can take $A=$ "", $B=$ "bca"), and a sequence of the first 7 beads (we can take $A=$ "b", $B=$ "ca").

In the second sample test, for example, a sequence of the first 13 beads is regular, if we take $A=$ "aba", $B=$ "ba".
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