codeforces#P1039A. Timetable

    ID: 29562 远端评测题 2000ms 512MiB 尝试: 0 已通过: 0 难度: 8 上传者: 标签>constructive algorithmsdata structuresgreedymath*2300

Timetable

Description

There are two bus stops denoted A and B, and there $n$ buses that go from A to B every day. The shortest path from A to B takes $t$ units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.

At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as $a_1 < a_2 < \ldots < a_n$ for stop A and as $b_1 < b_2 < \ldots < b_n$ for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least $t$ units of time later than departs.

It is known that for an order to be valid the latest possible arrival for the bus that departs at $a_i$ is $b_{x_i}$, i.e. $x_i$-th in the timetable. In other words, for each $i$ there exists such a valid order of arrivals that the bus departed $i$-th arrives $x_i$-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the $i$-th departed bus arrives $(x_i + 1)$-th.

Formally, let's call a permutation $p_1, p_2, \ldots, p_n$ valid, if $b_{p_i} \ge a_i + t$ for all $i$. Then $x_i$ is the maximum value of $p_i$ among all valid permutations.

You are given the sequences $a_1, a_2, \ldots, a_n$ and $x_1, x_2, \ldots, x_n$, but not the arrival timetable. Find out any suitable timetable for stop B $b_1, b_2, \ldots, b_n$ or determine that there is no such timetable.

The first line of the input contains two integers $n$ and $t$ ($1 \leq n \leq 200\,000$, $1 \leq t \leq 10^{18}$) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_1 < a_2 < \ldots < a_n \leq 10^{18}$), defining the moments of time when the buses leave stop A.

The third line contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \leq x_i \leq n$), the $i$-th of them stands for the maximum possible timetable position, at which the $i$-th bus leaving stop A can arrive at stop B.

If a solution exists, print "Yes" (without quotes) in the first line of the output.

In the second line print $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_1 < b_2 < \ldots < b_n \leq 3 \cdot 10^{18}$). We can show that if there exists any solution, there exists a solution that satisfies such constraints on $b_i$. If there are multiple valid answers you can print any of them.

If there is no valid timetable, print "No" (without quotes) in the only line of the output.

Input

The first line of the input contains two integers $n$ and $t$ ($1 \leq n \leq 200\,000$, $1 \leq t \leq 10^{18}$) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_1 < a_2 < \ldots < a_n \leq 10^{18}$), defining the moments of time when the buses leave stop A.

The third line contains $n$ integers $x_1, x_2, \ldots, x_n$ ($1 \leq x_i \leq n$), the $i$-th of them stands for the maximum possible timetable position, at which the $i$-th bus leaving stop A can arrive at stop B.

Output

If a solution exists, print "Yes" (without quotes) in the first line of the output.

In the second line print $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_1 < b_2 < \ldots < b_n \leq 3 \cdot 10^{18}$). We can show that if there exists any solution, there exists a solution that satisfies such constraints on $b_i$. If there are multiple valid answers you can print any of them.

If there is no valid timetable, print "No" (without quotes) in the only line of the output.

Samples

3 10
4 6 8
2 2 3

Yes
16 17 21 

2 1
1 2
2 1

No

Note

Consider the first example and the timetable $b_1, b_2, \ldots, b_n$ from the output.

To get $x_1 = 2$ the buses can arrive in the order $(2, 1, 3)$. To get $x_2 = 2$ and $x_3 = 3$ the buses can arrive in the order $(1, 2, 3)$. $x_1$ is not $3$, because the permutations $(3, 1, 2)$ and $(3, 2, 1)$ (all in which the $1$-st bus arrives $3$-rd) are not valid (sube buses arrive too early), $x_2$ is not $3$ because of similar reasons.