D. Range = √Sum

D. Range = √Sum

You are given an integer $n$. Find a sequence of $n$ distinct integers $a_1, a_2, \dots, a_n$ such that $1 \leq a_i \leq 10^9$ for all $i$ and $$\max(a_1, a_2, \dots, a_n) - \min(a_1, a_2, \dots, a_n)= \sqrt{a_1 + a_2 + \dots + a_n}.$$

It can be proven that there exists a sequence of distinct integers that satisfies all the conditions above.

Input

The first line of input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first and only line of each test case contains one integer $n$ ($2 \leq n \leq 3 \cdot 10^5$) — the length of the sequence you have to find.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output

For each test case, output $n$ space-separated distinct integers $a_1, a_2, \dots, a_n$ satisfying the conditions in the statement.

If there are several possible answers, you can output any of them. Please remember that your integers must be distinct!

Example

input

3
2
5
4

output

3 1
20 29 18 26 28
25 21 23 31

Note

In the first test case, the maximum is $3$, the minimum is $1$, the sum is $4$, and $3 - 1 = \sqrt{4}$.

In the second test case, the maximum is $29$, the minimum is $18$, the sum is $121$, and $29-18 = \sqrt{121}$.

For each test case, the integers are all distinct.

 

解题思路

　　构造题感觉更多是看直觉和灵感的，这里就直接按照题解的思路来写了。

　　对于$n$是奇数的情况，我们可以考虑以$n$为中心往两边构造出序列$$\left\{ {n - \left\lfloor {\frac{n}{2}} \right\rfloor,\, n - \left\lfloor {\frac{n}{2}} \right\rfloor - 1,\, \ldots \,,\, n-1,\, n,\, n+1,\, \ldots \,,\, n + \left\lfloor {\frac{n}{2}} \right\rfloor - 1,\, n + \left\lfloor {\frac{n}{2}} \right\rfloor} \right\}$$

　　此时序列的$\max - \min = n - 1$，总和为（两端两两相加）$\left\lfloor {\frac{n}{2}} \right\rfloor \times 2n + n = \frac{n-1}{2} \times 2n + n = n^2$。此时如果给序列中的每个数都加上$2$，那么$\max - \min = n - 1$不变，总和变成了${\frac{n-1}{2}} \times 2(n+2) + (n+2) = n^2 + 2n$。现在我们对序列中的最小值减去$1$，最大值加上$1$，此时$\max - \min = n + 1$，总和依然是$n^2 + 2n$。

　　为了使得$(n+1)^2 = n^2 + 2n + 1$，我们再对序列中第二大的数加上$1$就可以了（由于上一步已经让最大数加$1$了，因此对第二大数加$1$得到的结果是合法的）。

　　例如当$n = 5$：

• $[3, 4, 5, 6, 7]$（以$5$为中心构造）
• $[5, 6, 7, 8, 9]$（所有元素加$2$）
• $[4, 6, 7, 8, 10]$（最小值减$1$，最大值加$1$）
• $[4, 6, 7, 9, 10]$（第二大值加$1$）

　　对于$n$是偶数的情况，同样考虑以$n$为中心（不包含$n$）往两边构造出序列$$\left\{ {n - \frac{n}{2},\, n - \frac{n}{2} - 1,\, \ldots \,,\, n-1,\, n+1,\, \ldots \,,\, n + {\frac{n}{2}} - 1,\, n + {\frac{n}{2}} } \right\}$$

　　此时序列的$\max - \min = n$，总和为$\frac{n}{2} \times 2n = n^2$。

　　AC代码如下：

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

const int N = 3e5 + 10;

int a[N];

void solve() {
    int n;
    scanf("%d", &n);
    if (n & 1) {
        for (int i = 0, j = n - n / 2; i < n; i++) {
            a[i] = j++ + 2;
        }
        a[0]--, a[n - 1]++, a[n - 2]++;
    }
    else {
        for (int i = 0, j = n - n / 2; i < n; i++) {
            a[i] = j++;
            if (j == n) j++;
        }
    }
    for (int i = 0; i < n; i++) {
        printf("%d ", a[i]);
    }
    printf("\n");
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

　　cf的构造题还是经常无从下手，寒假需要多做这样的题进行训练。现在也准备期末了，没时间打cf和写题解了，寒假再开始吧（我真的好菜）｡･ﾟﾟ*(>д<)*ﾟﾟ･｡

 

参考资料

　　Codeforces Round #836 (Div. 2) Editorial：https://codeforces.com/blog/entry/109438