Cyclic Shifts Sorting

You are given an array a consisting of $n$ integers.

In one move, you can choose some index $i (1≤i≤n−2)$ and shift the segment $[a_i,a_{i+1},a_{i+2}]$ cyclically to the right (i.e. replace the segment $[a_i,a_{i+1},a_{i+2}]$ with $[a_{i+2},a_i,a_{i+1}]$).

Your task is to sort the initial array by no more than $n_2$ such operations or say that it is impossible to do that.

You have to answer t independent test cases.

Input
The first line of the input contains one integer $t (1≤t≤100)$ — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer $n (3≤n≤500)$ — the length of a. The second line of the test case contains n integers $a_1,a_2,…,a_n (1≤a_i≤500)$, where $a_i$ is the $i$-th element a.

It is guaranteed that the sum of n does not exceed $500$.

Output
For each test case, print the answer: $-1$ on the only line if it is impossible to sort the given array using operations described in the problem statement, or the number of operations ans on the first line and $ans$ integers $idx_1,idx_2,…,idx_{ans}$ $(1≤idx_i≤n−2)$, where $idx_i$ is the index of left border of the segment for the $i$-th operation. You should print indices in order of performing operations.

Example

input
5
5
1 2 3 4 5
5
5 4 3 2 1
8
8 4 5 2 3 6 7 3
7
5 2 1 6 4 7 3
6
1 2 3 3 6 4
output
0

6
3 1 3 2 2 3 
13
2 1 1 6 4 2 4 3 3 4 4 6 6 
-1
4
3 3 4 4 

代碼先貼上，題解坑留到考完試後再填

#include<bits/stdc++.h>

#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scahf("%c",&a);
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pc(a) putchar(a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define pii pair<int,int>
#define mii unordered_map<int,int>
#define msi unordered_map<string,int>
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define pr(x) cout<<#x<<": "<<x<<endl
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 505;
int T, n, m;
vi ans, seq;
int a[N];
int now;

inline void move(int x) {
    int pos;
    now++;
    if (m - now == 1) {
        if (a[1] > a[2]) {
            if (m == 2) {
                puts("-1");
                return;
            }
            if (a[2] == seq[now - 2])ans.pb(now - 1);
            else {
                int j = -1;
                repd(i, now - 2, 1)if (seq[i] == seq[i - 1]) {
                        j = i + 2;
                        break;
                    }
                if (j == -1) {
                    puts("-1");
                    return;
                }
                if ((now & 1) == (j & 1)) {
                    for (int i = now - 2; i >= j; i -= 2)ans.pb(i);
                    for (int i = now - 1; i >= j + 1; i -= 2)ans.pb(i);
                    ans.pb(j - 2), ans.pb(j - 2);
                    ans.pb(j - 1), ans.pb(j - 1);
                    for (int i = j + 1; i <= now - 1; i += 2)ans.pb(i), ans.pb(i);
                    for (int i = j; i <= now - 2; i += 2)ans.pb(i), ans.pb(i);
                } else {
                    for (int i = now - 1; i >= j; i -= 2)ans.pb(i);
                    for (int i = now - 1; i >= j; i -= 2)ans.pb(i);
                    ans.pb(j - 2), ans.pb(j - 2), ans.pb(j - 1);
                    for (int i = j; i <= now - 1; i += 2)ans.pb(i), ans.pb(i);
                    for (int i = j - 1; i <= now - 2; i += 2)ans.pb(i), ans.pb(i);
                }
            }
        }
        pi(ans.size());
        for (auto it:ans)printf("%d ", it);
        puts("");
        return;
    }
    repi(i, 1, n)if (a[i] == x) {
            pos = i;
            break;
        }
    n--;
    repi(i, pos, n)a[i] = a[i + 1];
    for (int i = now + pos - 3; i >= now; i -= 2)ans.pb(i);
    if (!(pos & 1)) {
        ans.pb(now), ans.pb(now);
        swap(a[1], a[2]);
    }
}

int main() {
    T = qr();
    while (T--) {
        ans.clear(), seq.clear(), now = 0;
        n = qr(), m = n;
        repi(i, 1, n)a[i] = qr(), seq.pb(a[i]);
        if (n == 1) {
            puts("0\n");
            continue;
        }
        sort(seq.begin(), seq.end());
        repi(i, 0, m - 2)move(seq[i]);
    }
    return 0;
}