Problem: You are given one integer n (n > 1).

Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).

Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).

You have to answer t independent test cases.

If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.

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 only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.

Output

For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).

If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.

Example

Input


2
2
5


Output


2 1
2 1 5 3 4

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        Rd in = new Rd();
        PrintWriter out = new PrintWriter(System.out);
        int T = in.nextInt();
        for(int t=0; t<T; t++){
            int n = in.nextInt();
            if(n%2==0) {
                for (int i = n; i >= 1; i--) {
                    out.print(i + ((i == 1) ? "" : " "));
                }
            } else{
                for (int i = n; i >= 1; i--) {
                    if(i== n/2){
                        out.print(i + 1 + " ");
                    }else if(i==n/2+1){
                        out.print(i - 1 + " ");
                    }
                    else out.print(i + ((i == 1) ? "" : " "));
                }
            }
            out.println();

        }
        out.flush();

    }

    static class Rd {
        BufferedReader br;
        StringTokenizer st;

        public Rd() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    throw null;
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        int[] nextIntArray(int n) {
            int r[] = new int[n];
            for (int i = 0; i < n; i++) r[i] = nextInt();
            return r;
        }

        long[] nextLongArray(int n) {
            long r[] = new long[n];
            for (int i = 0; i < n; i++) r[i] = nextLong();
            return r;
        }

        char[][] grid(int r, int c) {
            char res[][] = new char[r][c];
            for (int i = 0; i < r; i++) {
                char l[] = next().toCharArray();
                for (int j = 0; j < c; j++) {
                    res[i][j] = l[j];
                }
            }
            return res;
        }
    }

}