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:
// /*
// power code taken from geeks for geeks
// */
// cout << setprecision (2) << fixed << 1.2;

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;

 
// THINGS TO REMEMBER
// ENDL is slow, '\n' is fast
// Clear everything (including graphs) between test cases
// use anti-anti-hash: https://codeforces.com/blog/entry/62393
// pb-ds: https://codeforces.com/blog/entry/11080
// check when to use LLONG_MAX/LLONG_MIN vs INT_MAX or INT_MIN
// You frequently suffer from confirmation bias - you trust your initial solution and miss simple things. 
// When you hit a roadblock, remember to rethink the solution ground up, not just try hacky fixes
// do something instead of nothing and stay organized


#define PLL pair<ll, ll>
const ll INF = 1e18 + 5;
const ll MAX_N = 3e2 + 5;
const ll MODD = 1e9 + 7;
const ll H = 20LL;

ll max(ll a,ll b){return a>b?a:b;}
ll min(ll a,ll b){return a<b?a:b;}
double max(double a,double b){return a>b?a:b;}
double min(double a,double b){return a<b?a:b;}

ll modd(ll a){if(a<0LL) return a*-1LL;return a;}
double modd(double a){if(a<0.0) return a*-1.0;return a;}

ll modulo(ll a){return (a + MODD) % MODD; }

ll power(ll k, ll p){
    if (p == 0)
        return 1LL;
 
    ll P = power(k, p/2LL);
    P = (P * P) % MODD;
 
    if (p % 2 == 1)
        P = (P * k) % MODD;
 
    return P;
}



void solve() {
    ll n;
    cin >> n;
    vector<ll> ans;
    for(ll i = 1; i <= n; i++) {
        ans.push_back(i);
    }
    for(ll i = 0; i + 1 < n; i += 2) {
        swap(ans[i], ans[i + 1]);
    }
    if(n % 2 == 1) {
        swap(ans[n - 1], ans[n - 2]);
    }
    for(ll i: ans) {
        cout << i << " ";
    }
    cout << endl;
}

int main(){


    #ifndef ONLINE_JUDGE 

    // For getting input from input.txt file 
    freopen("input.txt", "r", stdin); 

    // Printing the Output to output.txt file 
    freopen("output.txt", "w", stdout); 
    #endif 


    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    



    ll t=1;


    cin >> t;
    

    for(ll i = 0; i < t; i++) {
        
        // cout << "---------------case " << i + 1 << endl;

        // cout << "Case #" << i + 1 << ": ";
        solve();
    }


    return 0;
}