CF377B.Preparing for the Contest

Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get c__i 'passes' in his subjects (regardless of the volume of his work).

Bugs, like students, are not the same: every bug is characterized by complexity a__j, and every student has the level of his abilities b__i. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: b__i ≥ a__j, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.

The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.

很快将举办全球规模最大的编程竞赛，但评测系统目前仍有 $m$ 个缺陷（bug）。赛事主办方——一所知名大学——别无选择，只能招募本校学生来修复所有缺陷。该校共有 $n$ 名学生能够胜任此项工作。学生们意识到自己是主办方唯一的希望，因此不愿无偿劳动：第 $i$ 名学生要求获得 $c_i$ 张课程“免修券”（pass），且该要求与其实际工作量无关。

缺陷与学生一样各不相同：每个缺陷 $j$ 具有复杂度 $a_j$，而每名学生 $i$ 具备能力水平 $b_i$。学生 $i$ 仅当其能力水平不低于缺陷 $j$ 的复杂度（即 $b_i \ge a_j$）时，才能修复该缺陷，且需耗时一天；否则该缺陷必须交由其他学生修复。当然，任何学生在一天内至多只能修复一个缺陷。所有缺陷彼此独立，因此可按任意顺序修复，且不同学生可并行工作。

该校希望在满足向学生发放的免修券总数不超过 $s$ 张的前提下，尽快修复全部缺陷。请确定应选用哪些学生，并制定一份详细的工作计划，指明每名学生具体负责修复哪个缺陷。

The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.

The next line contains m space-separated integers _a_1, _a_2, ..., a__m (1 ≤ a__i ≤ 109) — the bugs' complexities.

The next line contains n space-separated integers _b_1, _b_2, ..., b__n (1 ≤ b__i ≤ 109) — the levels of the students' abilities.

The next line contains n space-separated integers _c_1, _c_2, ..., c__n (0 ≤ c__i ≤ 109) — the numbers of the passes the students want to get for their help.

第一行包含三个用空格分隔的整数：$n$、$m$ 和 $s$（$1 \leq n, m \leq 10^5$，$0 \leq s \leq 10^9$）—— 分别表示学生人数、系统中缺陷（bugs）的数量，以及学校愿意向学生发放的通行证（passes）的最大数量。

第二行包含 $m$ 个用空格分隔的整数 $a_1, a_2, \dots, a_m$（$1 \leq a_i \leq 10^9$）—— 表示各缺陷的复杂度。

第三行包含 $n$ 个用空格分隔的整数 $b_1, b_2, \dots, b_n$（$1 \leq b_i \leq 10^9$）—— 表示各学生的能力等级。

第四行包含 $n$ 个用空格分隔的整数 $c_1, c_2, \dots, c_n$（$0 \leq c_i \leq 10^9$）—— 表示各学生为其协助所要求的通行证数量。

If the university can't correct all bugs print "NO".

Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.

如果大学无法修复所有漏洞，则输出“NO”。

否则，第一行输出“YES”，第二行输出 m 个以空格分隔的整数：其中第 i 个数应表示在最优方案中修复第 i 个漏洞的学生编号。漏洞应尽可能快地被修复（即所用天数必须最少），且总共发放的通过次数不得超过 s。若存在多个最优解，输出任意一个即可。

Consider the first sample.

The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).

The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.

考虑第一个样例。

第三位学生（能力等级为 3）必须修复第 2 个和第 4 个缺陷（其复杂度分别为 3 和 2），第二位学生（能力等级为 1）必须修复第 1 个和第 3 个缺陷（它们的复杂度也均为 1）。每位学生修复每个缺陷均需一天，因此修复所有缺陷共需 2 天（学生可以并行工作）。

第二位学生要求 3 次协助机会，第三位学生要求 6 次协助机会。这符合大学的能力限制，因为大学最多可提供 9 次协助机会。