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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define mod 1000000007
int main() {
    ll t;
    cin >> t;
    while(t--) {
        ll n;
        cin >> n;
        ll arr[n];
        ll c = 0; 
        for(ll i = n-1; i >= 0; i--) {
            arr[i] = c+1;
            c++;
        }
        if(n%2 != 0) {
            ll temp = arr[n/2];
            arr[n/2] = arr[n/2 + 1];
            arr[n/2 + 1] = temp;
        }
        for(ll i = 0; i < n; i++) {
            cout << arr[i] << " ";
        }
        cout << endl;
    }
    
}