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.util.*;
import java.lang.Math;
import java.util.concurrent.TimeUnit;
import java.util.stream.Collectors;
import java.io.*;
import java.math.BigInteger;
import java.io.FileWriter;

public class CodeForces1 {

	static class FastReader

	{
		BufferedReader br;
		StringTokenizer st;

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

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

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

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

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static FastReader f = new FastReader();

	static void print(String s) {
		System.out.print(s);
	}

	static void print(boolean s) {
		System.out.print(s);
	}

	static boolean isPrime(long n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		// This is checked so that we can skip
		// middle five numbers in below loop
		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	static long[] inArr(long a[], int n) {
		for (int i = 0; i < n; i++)
			a[i] = f.nextLong();
		return (a);
	}

	static void printSubsets(int set[]) {
		int n = set.length;
		Arrays.sort(set);
		for (int i = 0; i < n; i++) {

		}
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int i = 0; i < (1 << n); i++) {
			for (int j = 0; j < n; j++) {
				if ((i & (1 << j)) > 0) {
					if (hm.get(set[j]) == null) {
						hm.put(set[j], 0);
					}
					hm.put(set[j], hm.get(set[j]) + 1);

				}
			}

		}
	}

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	static long power(long x, long y, long p) {

		// Initialize result
		long res = 1;

		// Update x if it is more than or
		// equal to p
		x = x % p;

		while (y > 0) {

			// If y is odd, multiply x
			// with result
			if (y % 2 == 1)
				res = (res * x) % p;

			// y must be even now
			y = y >> 1; // y = y/2
			x = (x * x) % p;
		}

		return res;
	}

	static long modInverse(long n, long p) {
		return power(n, p - 2, p);
	}

	static long nCrModPFermat(int n, int r) {
		long p = 1000000007;

		// Base case
		if (r == 0)
			return 1;

		// Fill factorial array so that we
		// can find all factorial of r, n
		// and n-r
		long[] fac = new long[n + 1];
		fac[0] = 1;

		for (int i = 1; i <= n; i++)
			fac[i] = fac[i - 1] * i % p;

		return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p;
	}
	static int lcm(int a, int b) 
    { 
        return (a / __gcd(a, b) * b); 
    } 
  
    static int __gcd(int a, int b)  
    { 
        return b == 0 ? a : __gcd(b, a % b); 
    } 
  
    // Function to find and print the two numbers 
    static int[] findNums(int x) 
    { 
  
        int ans = -1; 
  
        // To find the factors 
        for (int i = 1; i <= Math.sqrt(x); i++) 
        { 
  
            // To check if i is a factor of x and 
            // the minimum possible number 
            // satisfying the given conditions 
            if (x % i == 0 && lcm(i, x / i) == x) 
            { 
  
                ans = i; 
            } 
		} 
		int a[]={ans,x/ans};
        return (a);//(ans + " " + (x / ans)); 
    }
	public static void main(String args[]) throws Exception {
		BufferedWriter w = new BufferedWriter(new OutputStreamWriter(System.out));
		Random rand = new Random();
		int t = f.nextInt();
		while (t-- != 0) 
		{
			int n=f.nextInt();
			if(n%2==1)
			{
				w.write((n/2+1)+" ");
			}
			for(int i=n;i>=1;i--)
			{
				if(n%2==0)
					w.write(i+" ");
				else
				{
					if(i!=n/2+1)
					{
						w.write(i+" ");
					}
				}
			}
			w.write("\n");
			
		}
		w.flush();

	}
}