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:
#include <bits/stdc++.h>
#define forn(i, n) for(int i = 0; i < (n); i++)
#define forab(i, a, b) for(int i = (a); i <= (b); i++)
#define forba(i, b, a) for(int i = (b); i >= (a); i--)
#define INF 0x3f3f3f3f
#define mk(x) (1<<(x)) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '\n'
#define lowbit(x) (x&-x)

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

template<class T> bool ckmin(T& a, const T& b) { return a > b ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {if (ch == '-') fg = -1;ch = getchar();}
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar('-'), x = -x;
    do{++len; c[len] = x%10 + '0';} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

#ifdef ONLINE_JUDGE
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#else
#define fast 
#define N 105
#define M 105
#endif

int n;
int a[N];
deque<pair<int, int>> ops;
ll ans;
inline void add(int l, int r, bool isfront) {
    if (isfront)
        ops.push_front({l, r});
    else
        ops.push_back({l, r});
    ans += (ll)(r-l) * (r-l);
}
int solveTestCase() {
    cin >> n;
    forab(i, 1, n) a[i] = i;
    ans = 0;
    ops.clear();
    int l = 1, r = n;
    add(l, r, false);
    for (; sz(ops) < n-1; l++, r--) {
        if (sz(ops) < n-1) {
            add(l+1, n, true);
        }
        if (sz(ops) < n-1) {
            add(1, r-1, true);
        }
    }
    for (int i = 1; i <= n-1; i++) {
        auto tmp = ops.back();
        ops.pop_back();
        ops.push_front(tmp);
        swap(a[tmp.first], a[tmp.second]);
    }
    cout << ans << endl;
    forab(i, 1, n) {
        cout << a[i];
        if (i == n) cout << endl;
        else cout << ' ';
    }
    cout << n-1 << endl;
    forab(i, 1, n-1) {
        auto tmp = ops.front(); ops.pop_front();
        if (a[tmp.second] == tmp.first)
            cout << tmp.first << ' ' << tmp.second << endl;
        else
            cout << tmp.second << ' ' << tmp.first << endl;
        swap(a[tmp.first], a[tmp.second]);
    }
    return 0;
}

int main() {
    fast;
    int t = 1;
    cin >> t;
    while (t--)
        solveTestCase();
    return 0;
}
/*

*/