CF1039A.Timetable

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 $a_1 \lt a_2 \lt \ldots \lt a_n$ for stop A and as $b_1 \lt b_2 \lt \ldots \lt b_n$ 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.

有两个公交站，分别记为 A 和 B，每天有 $n$ 辆公交车从 A 开往 B。从 A 到 B 的最短路径耗时 $t$ 个单位时间，但某些公交车可能选择更长的路径。此外，公交车在行驶途中允许相互超车。

在每个车站，均可查到一份按时间顺序排列的公交车到站时刻列表。我们将该列表在 A 站记为 $a_1 \lt a_2 \lt \ldots \lt a_n$，在 B 站记为 $b_1 \lt b_2 \lt \ldots \lt b_n$。公交车始终严格依照时刻表从 A 出发、在 B 到达，但到达 B 的顺序可能与出发顺序不同。我们称一个到达顺序是有效的，当且仅当每辆公交车的到达时刻至少比其出发时刻晚 $t$ 个单位时间。

已知：对于每个 $i$，若某辆公交车在 A 站第 $i$ 个出发（即出发时刻为 $a_i$），则其在 B 站最晚可能的到达位置（即在 B 站时刻表 $b_1, b_2, \ldots, b_n$ 中的序号）为 $b_{x_i}$。换言之，对每个 $i$，存在某个有效的到达顺序，使得第 $i$ 个出发的公交车恰好是第 $x_i$ 个到达的（其余公交车的到达次序可任意）；但不存在任何有效的到达顺序，使得第 $i$ 个出发的公交车是第 $(x_i + 1)$ 个到达的。

形式化地，称一个排列 $p_1, p_2, \ldots, p_n$ 是有效的，当且仅当对所有 $i$ 均满足 $b_{p_i} \ge a_i + t$。那么 $x_i$ 即为所有有效排列中 $p_i$ 的最大可能取值。

现给定序列 $a_1, a_2, \ldots, a_n$ 和 $x_1, x_2, \ldots, x_n$，但未给出 B 站的到达时刻表。请构造出任意一组满足条件的 B 站时刻表 $b_1, b_2, \ldots, b_n$，或判定不存在这样的时刻表。

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 $a_1, a_2, \ldots, a_n$ ($1 \leq a_1 \lt a_2 \lt \ldots \lt a_n \leq 10^{18}$), 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.

输入的第一行包含两个整数 $n$ 和 $t$（$1 \leq n \leq 200\,000$，$1 \leq t \leq 10^{18}$）——分别表示时刻表中公交车的数量，以及从站点 A 到站点 B 的最小可能旅行时间。

第二行包含 $n$ 个整数 $a_1, a_2, \ldots, a_n$（$1 \leq a_1 \lt a_2 \lt \ldots \lt a_n \leq 10^{18}$），表示各辆公交车离开站点 A 的时刻。

第三行包含 $n$ 个整数 $x_1, x_2, \ldots, x_n$（$1 \leq x_i \leq n$），其中第 $i$ 个数表示：第 $i$ 辆从站点 A 出发的公交车，最迟可于时刻表中的第 $x_i$ 个位置到达站点 B。

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

In the second line print $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_1 \lt b_2 \lt \ldots \lt b_n \leq 3 \cdot 10^{18}$). 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.

如果存在解，则在输出的第一行打印“Yes”（不带引号）。

在第二行打印 $n$ 个整数 $b_1, b_2, \ldots, b_n$（满足 $1 \leq b_1 \lt b_2 \lt \ldots \lt b_n \leq 3 \cdot 10^{18}$）。我们可以证明：若存在任意解，则必存在一个满足上述关于 $b_i$ 的约束条件的解。若存在多个合法答案，可输出其中任意一个。

若不存在合法的时间表，则在输出的唯一一行中打印“No”（不带引号）。

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.

考虑第一个样例以及输出中的时刻表 $b_1, b_2, \ldots, b_n$。

为使 $x_1 = 2$，公交车到达顺序可以是 $(2, 1, 3)$；为使 $x_2 = 2$ 且 $x_3 = 3$，公交车到达顺序可以是 $(1, 2, 3)$。$x_1$ 不可能是 $3$，因为排列 $(3, 1, 2)$ 和 $(3, 2, 1)$（即第 $1$ 辆公交车第 $3$ 个到达的所有情况）均不合法（某些公交车到达过早）；同理，$x_2$ 也不可能是 $3$。