#AT2393. E - Cheating Amidakuji

E - Cheating Amidakuji

E - Cheating Amidakuji

Score : $500$ points

Problem Statement

Let us trace the path from the top of vertical bar 11, and let the score be the number of the vertical bar we reach at the bottom.

For each i=1,2,,Mi=1,2,\dots,M, answer the following question.

  • Let SiS_i be the score assuming that horizontal bar ii is erased. Find SiS_i.

Note that horizontal bars are not actually erased, so the questions are independent of each other.

Below is a more formal description of the problem. -->

You are given a sequence of length $M$ consisting of integers between $1$ and $N-1$, inclusive: $A=(A_1,A_2,\dots,A_M)$. Answer the following question for $i=1, 2, \dots, M$.

  • There is a sequence $B=(B_1,B_2,\dots,B_N)$. Initially, we have $B_j=j$ for each $j$. Let us perform the following operation for $k=1, 2, \dots, i-1, i+1, \dots, M$ in this order (in other words, for integers $k$ between $1$ and $M$ except $i$ in ascending order).
    • Swap the values of $B_{A_k}$ and $B_{A_k+1}$.
  • After all the operations, let $S_i$ be the value of $j$ such that $B_j=1$. Find $S_i$.

Constraints

  • $2 \leq N \leq 2\times 10^5$
  • $1 \leq M \leq 2\times 10^5$
  • $1 \leq A_i \leq N-1\ (1\leq i \leq M)$
  • All values in the input are integers.

Input

The input is given from Standard Input in the following format:

NN MM

A1A_1 A2A_2 \dots AMA_M

Output

Print $M$ lines. The $i$-th line $(1\leq i \leq M)$ should contain the value $S_i$ as an integer.


5 4
1 2 3 2
1
3
2
4

For $i = 2$, the operations change $B$ as follows.

  • Initially, $B = (1,2,3,4,5)$.
  • Perform the operation for $k=1$. That is, swap the values of $B_1$ and $B_2$, making $B = (2,1,3,4,5)$.
  • Perform the operation for $k=3$. That is, swap the values of $B_3$ and $B_4$, making $B = (2,1,4,3,5)$.
  • Perform the operation for $k=4$. That is, swap the values of $B_2$ and $B_3$, making $B = (2,4,1,3,5)$.

After all the operations, we have $B_3=1$, so $S_2 = 3$.

Similarly, we have the following.

  • For $i=1$: performing the operation for $k=2,3,4$ in this order makes $B=(1,4,3,2,5)$, so $S_1=1$.
  • For $i=3$: performing the operation for $k=1,2,4$ in this order makes $B=(2,1,3,4,5)$, so $S_3=2$.
  • For $i=4$: performing the operation for $k=1,2,3$ in this order makes $B=(2,3,4,1,5)$, so $S_4=4$.

3 3
2 2 2
1
1
1

10 10
1 1 1 9 4 4 2 1 3 3
2
2
2
3
3
3
1
3
4
4