Problem:
Score : 600 points

Problem Statement

Takahashi is playing a board game called Sugoroku.

On the board, there are N + 1 squares numbered 0 to N. Takahashi starts at Square 0, and he has to stop exactly at Square N to win the game.

The game uses a roulette with the M numbers from 1 to M. In each turn, Takahashi spins the roulette. If the number x comes up when he is at Square s, he moves to Square s+x. If this makes him go beyond Square N, he loses the game.

Additionally, some of the squares are Game Over Squares. He also loses the game if he stops at one of those squares. You are given a string S of length N + 1, representing which squares are Game Over Squares. For each i (0 \leq i \leq N), Square i is a Game Over Square if S[i] = 1 and not if S[i] = 0.

Find the sequence of numbers coming up in the roulette in which Takahashi can win the game in the fewest number of turns possible. If there are multiple such sequences, find the lexicographically smallest such sequence. If Takahashi cannot win the game, print -1.

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^5

|S| = N + 1

S consists of 0 and 1.

S[0] = 0

S[N] = 0

Input

Input is given from Standard Input in the following format:

N M
S

Output

If Takahashi can win the game, print the lexicographically smallest sequence among the shortest sequences of numbers coming up in the roulette in which Takahashi can win the game, with spaces in between.

If Takahashi cannot win the game, print -1.

Sample Input 1

9 3
0001000100

Sample Output 1

1 3 2 3

If the numbers 1, 3, 2, 3 come up in this order, Takahashi can reach Square 9 via Square 1, 4, and 6. He cannot reach Square 9 in three or fewer turns, and this is the lexicographically smallest sequence in which he reaches Square 9 in four turns.

Sample Input 2

5 4
011110

Sample Output 2

-1

Takahashi cannot reach Square 5.

Sample Input 3

6 6
0101010

Sample Output 3

6

Solution in PyPy3 (7.3.0) (judge verdict: Accepted)

N, M = map(int, input().split())
S = input()[::-1]

"""
スタートから貪欲に考えようとすると
「最短で行くために大きい目を出したい」と
「辞書順最小のために小さい目を出したい」が共存してやりづらい

ゴールから貪欲に考える（出来るだけ多く戻る）と考えると
「最短で行くために大きい目を出したい」と
「辞書順最小のために、最後の手を出来るだけ小さくしたい
 -> 後ろから考えて、大きい目を出せば良い」となり
貪欲に解ける
"""
# 出した目の値（逆順）
ans = []
# 現在の位置（ゴールを0、スタートをNとする）
now = 0

while True:
    # Mだとオーバーしてしまう場合(ちょうど終わる場合も含む)
    # 最後の目を記録して終了
    if now + M >= N:
        ans.append(N - now)
        break

    # Mマス進むと仮定する
    step = M
    # 行き先がゲームオーバーなら、stepをデクリメント
    while S[now + step] == '1':
        step -= 1
        # step = M,M-1,...,1がダメならこれ以上進めないので終了
        if step == 0:
            print(-1)
            exit()

    now += step
    ans.append(step)

print(*ans[::-1])