欢迎来到起遇信息学
起遇信息学正处于上线筹建阶段,以下功能已全部开放免费体验: ✅ 完整题库浏览与代码提交评测(C / C++ / Python / Java 等) ✅ 入门到进阶的系列课程试读、作业与考试 ✅ AI 提示、AI 作业分析等智能助教功能 ✅ 赛事模拟与个人能力报告 ✅ 邮箱注册开放 ⏳ 付费课程订阅与微信/支付宝支付通道 ⏳ 手机号登录,微信扫码登录、微信公众号绑定 使用中如遇任何问题,欢迎通过页面底部 **"联系我们"** 与我们沟通。
CF1823B.Sort with Step
Sort with Step
Let's define a permutation of length as an array of length , which contains every number from to exactly once.
You are given a permutation and a number . You need to sort this permutation in the ascending order. In order to do it, you can repeat the following operation any number of times (possibly, zero):
- pick two elements of the permutation and such that , and swap them.
Unfortunately, some permutations can't be sorted with some fixed numbers . For example, it's impossible to sort with .
That's why, before starting the sorting, you can make at most one preliminary exchange:
- choose any pair and and swap them.
Your task is to:
- check whether is it possible to sort the permutation without any preliminary exchanges,
- if it's not, check, whether is it possible to sort the permutation using exactly one preliminary exchange.
For example, if and permutation is , then you can make a preliminary exchange of and , which will produce permutation , which is possible to sort with given .
Input
Each test contains multiple test cases. The first line contains the number of test cases (). The description of the test cases follows.
The first line of each test case contains two integers and (; ) — length of the permutation, and a distance between elements that can be swapped.
The second line of each test case contains integers () — elements of the permutation .
It is guaranteed that the sum of over all test cases does not exceed .
Output
For each test case print
0, if it is possible to sort the permutation without preliminary exchange;1, if it is possible to sort the permutation with one preliminary exchange, but not possible without preliminary exchange;-1, if it is not possible to sort the permutation with at most one preliminary exchange.
Note
In the first test case, there is no need in preliminary exchange, as it is possible to swap and then .
In the second test case, there is no need in preliminary exchange, as it is possible to swap and then .
In the third test case, you need to apply preliminary exchange to . After that the permutation becomes and can be sorted with .
Samples
6
4 1
3 1 2 4
4 2
3 4 1 2
4 2
3 1 4 2
10 3
4 5 9 1 8 6 10 2 3 7
10 3
4 6 9 1 8 5 10 2 3 7
10 3
4 6 9 1 8 5 10 3 2 7
0
0
1
0
1
-1
在线编程 IDE
建议全屏模式获得最佳体验
| 进入全屏编程 | Alt+E |
| 递交评测 | Ctrl+Enter |
| 注释/取消注释 | Ctrl+/ |
| 缩放字体 | Ctrl+滚轮 |