Problem: Lunar rover finally reached planet X. After landing, he met an obstacle, that contains permutation p of length n. Scientists found out, that to overcome an obstacle, the robot should make p an identity permutation (make p_i = i for all i).

Unfortunately, scientists can't control the robot. Thus the only way to make p an identity permutation is applying the following operation to p multiple times: 

  * Select two indices i and j (i ≠ j), such that p_j = i and swap the values of p_i and p_j. It takes robot (j - i)^2 seconds to do this operation. 

Positions i and j are selected by the robot (scientists can't control it). He will apply this operation while p isn't an identity permutation. We can show that the robot will make no more than n operations regardless of the choice of i and j on each operation.

Scientists asked you to find out the maximum possible time it will take the robot to finish making p an identity permutation (i. e. worst-case scenario), so they can decide whether they should construct a new lunar rover or just rest and wait. They won't believe you without proof, so you should build an example of p and robot's operations that maximizes the answer.

For a better understanding of the statement, read the sample description.

Input

The first line of input contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Each of next t lines contains the single integer n (2 ≤ n ≤ 10^5) – the length of p.

Note, that p is not given to you. You should find the maximum possible time over all permutations of length n.

It is guaranteed, that the total sum of n over all test cases doesn't exceed 10^5.

Output

For each test case in the first line, print how many seconds will the robot spend in the worst case.

In the next line, print the initial value of p that you used to construct an answer.

In the next line, print the number of operations m ≤ n that the robot makes in your example.

In the each of next m lines print two integers i and j — indices of positions that the robot will swap on this operation. Note that p_j = i must holds (at the time of operation).

Example

Input


3
2
3
3


Output


1
2 1
1
2 1
5
2 3 1
2
1 3
3 2
5
2 3 1
2
1 3
2 3

Note

For n = 2, p can be either [1, 2] or [2, 1]. In the first case p is already identity, otherwise robot will make it an identity permutation in 1 second regardless of choise i and j on the first operation.

For n = 3, p can be equals [2, 3, 1].

  * If robot will select i = 3, j = 2 on the first operation, p will become [2, 1, 3] in one second. Now robot can select only i = 1, j = 2 or i = 2, j = 1. In both cases, p will become identity in one more second (2 seconds in total).
  * If robot will select i = 1, j = 3 on the first operation, p will become [1, 3, 2] in four seconds. Regardless of choise of i and j on the second operation, p will become identity in five seconds. 



We can show, that for permutation of length 3 robot will always finish all operation in no more than 5 seconds.

Solution (in CPP):

Solution:
/*
Author: Wa-Automaton
Time: 202101203 15:20:00 UTC+8       

Website: codeforces 
Contest: Codeforces Round #696 (Div. 2)
Problem: E. What Is It?

Result: 
Solutions: 
    
*/

#include<bits/stdc++.h>
using namespace std;

#define lson (root << 1)
#define rson (root << 1 | 1)

typedef long long LL;
typedef pair<LL, LL> PLL;
typedef vector<LL> VL;
typedef vector<PLL> VLL;
typedef vector<VL> VVL;

const LL MOD = 1e9 + 7;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
const LL Inf = 0x3f3f3f3f;
const LL N = 2e5 + 7;

LL n, a[N], ans;
VLL g;
// 
void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) a[i] = i;
    ans = 0; g.clear();
    for (LL l, r, i = 1; i < n; ++i) {
        l = i - 1;
        r = n - i;
        if (l >= r) {
            swap(a[i], a[1]);
            g.push_back({i, 1});
            ans += l * l;
        } else {
            swap(a[i], a[n]);
            g.push_back({i, n});
            ans += r * r;
        }
    }
    cout << ans << endl;
    for (int i = 1; i <= n; ++i) cout << a[i] << " "; cout << endl;
    cout << n - 1 << endl;
    for (int i = n - 2; i >= 0; --i) {
        cout << g[i].first << " " << g[i].second << endl;
    }
}

signed main()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) 
        solve();
    return 0;
}