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:
// package SpecialPermutation;

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) throws IOException {
        Reader.init(System.in);
        int T = Reader.nextInt();

        while (T-- != 0) {
            int N = Reader.nextInt();

            ans(N);

        }

    }

    private static void ans(int n) throws IOException {
        String s = "";

        for (int i = 0; i < n - 1; i++) {
            s += (i + 1) + " ";
        }
        s = n + " " + s;

        Reader.println(s);

    }
}

/** Class for buffered Maing int and double values */
class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;
    static BufferedWriter log;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(new InputStreamReader(input));
        tokenizer = new StringTokenizer("");
        log = new BufferedWriter(new OutputStreamWriter(System.out));
    }

    /** get next word */
    static String next() throws IOException {
        while (!tokenizer.hasMoreTokens()) {

            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

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

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

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

    static void print(String s) throws IOException {
        log.write(s);
        log.flush();
    }

    static void println(String s) throws IOException {
        log.write(s + "\n");
        log.flush();
    }
}