Codeforces Round #636 (Div. 3) D.Constant Palindrome Sum

题目链接
You are given an array a consisting of n integers (it is guaranteed that n is even, i.e. divisible by 2). All ai does not exceed some integer k.

Your task is to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace ai with some integer in range [1;k]) to satisfy the following conditions:

after all replacements, all ai are positive integers not greater than k;
for all i from 1 to n2 the following equation is true: ai+an−i+1=x, where x should be the same for all n2 pairs of elements.
You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤1e4) — the number of test cases. Then t test cases follow.

The first line of the test case contains two integers n and k (2≤n≤2e5,1≤k≤2e5) — the length of a and the maximum possible value of some ai correspondingly. It is guratanteed that n is even (i.e. divisible by 2). The second line of the test case contains n integers a1,a2,…,an (1≤ai≤k), where ai is the i-th element of a.

It is guaranteed that the sum of n (as well as the sum of k) over all test cases does not exceed 2⋅105 (∑n≤2e5, ∑k≤2e5).

Output

For each test case, print the answer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.

Example

input

4
4 2
1 2 1 2
4 3
1 2 2 1
8 7
6 1 1 7 6 3 4 6
6 6
5 2 6 1 3 4

output

比较巧妙的一道题目~
我们任意想到暴力，即从 [ 2 , 2 ∗ k ] [2,2*k] [2,2∗k] 里面一个个枚举，找最少替代次数，这样想肯定是会超时的，所以我们考虑用预处理的方式快速计算每个数需要的最少替代次数
首先用 c n t [ i ] cnt[i] cnt[i] 记录每一对 a [ i ] + a [ n − i − 1 ] a[i]+a[n-i-1] a[i]+a[n−i−1] 出现的次数
然后用 p r e [ i ] pre[i] pre[i] 记录每一对 a [ i ] , a [ n − i − 1 ] a[i],a[n-i-1] a[i],a[n−i−1] 变换一次得到的数的范围，很容易发现这个范围就是 [ m i n ( a [ i ] , a [ n − i − 1 ] ) + 1 , m a x ( a [ i ] , a [ n − i − 1 ] ) + k ] [min(a[i],a[n-i-1])+1,max(a[i],a[n-i-1])+k] [min(a[i],a[n−i−1])+1,max(a[i],a[n−i−1])+k]，通过记录端点的值控制区间修改，即 p r e [ m i n ( a [ i ] , a [ n − i − 1 ] ) + 1 ] + 1 pre[min(a[i],a[n-i-1])+1]+1 pre[min(a[i],a[n−i−1])+1]+1， p r e [ m a x ( a [ i ] , a [ n − i − 1 ] ) + k + 1 ] − 1 pre[max(a[i],a[n-i-1])+k+1]-1 pre[max(a[i],a[n−i−1])+k+1]−1，那么对每个数 i i i，所需的最小变换次数就是 ( n / 2 − p r e [ i ] ) ∗ 2 + p r e [ i ] − s u m [ i ] (n/2-pre[i])*2+pre[i]-sum[i] (n/2−pre[i])∗2+pre[i]−sum[i]，遍历更新一下答案即可，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;int main(){int t;cin>>t;while(t--){int n,k;cin>>n>>k;vector<int>a(n);for(auto &i:a) cin>>i;vector<int>cnt(2*k+1);for(int i=0;i<n/2;i++)cnt[a[i]+a[n-i-1]]++;vector<int>pre(2*k+2);for(int i=0;i<n/2;i++){int l1=a[i]+1,r1=a[i]+k;int l2=a[n-i-1]+1,r2=a[n-i-1]+k;pre[min(l1,l2)]++;pre[max(r1,r2)+1]--;}for(int i=1;i<=2*k+1;i++) pre[i]+=pre[i-1];int ans=1e9;for(int i=2;i<=2*k;i++){ans=min(ans,(pre[i]-cnt[i])+(n/2-pre[i])*2);}cout<<ans<<endl;}return 0;
}