C. MEX Repetition

C. MEX Repetition

You are given an array $a_1,a_2,\ldots, a_n$ of pairwise distinct integers from $0$ to $n$. Consider the following operation:

• consecutively for each $i$ from $1$ to $n$ in this order, replace $a_i$ with $\operatorname{MEX}(a_1, a_2, \ldots, a_n)$.

Here $\operatorname{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_m$ is defined as the smallest non-negative integer $x$ which does not occur in the collection $c$. For example, $\operatorname{MEX}(0, 2, 2, 1, 4) = 3$ and $\operatorname{MEX}(1, 2) = 0$.

Print the array after applying $k$ such operations.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($1\le n\le 10^5$, $1\le k\le 10^9$).

The second line contains $n$ pairwise distinct integers $a_1,a_2,\ldots, a_n$ ($0\le a_i\le n$) representing the elements of the array before applying the operations.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output

For each test case, print all $n$ elements of the array after applying $k$ operations.

Example

input

5
1 2
1
3 1
0 1 3
2 2
0 2
5 5
1 2 3 4 5
10 100
5 3 0 4 2 1 6 9 10 8

output

1
2 0 1
2 1
2 3 4 5 0
7 5 3 0 4 2 1 6 9 10

Note

In the first test case, here is the entire process:

1. On the first operation, the array changes from $[1]$ to $[0]$, since $\operatorname{MEX}(1) = 0$.
2. On the second operation, the array changes from $[0]$ to $[1]$, since $\operatorname{MEX}(0) = 1$.

Thus, the array becomes $[1]$ after two operations.

In the second test case, the array changes as follows during one operation: $[{\mkern3mu\underline{\mkern-3mu {\bf 0}\mkern-3mu}\mkern3mu}, 1, 3] \rightarrow [2, {\mkern3mu\underline{\mkern-3mu {\bf 1}\mkern-3mu}\mkern3mu}, 3] \rightarrow [2, 0, {\mkern3mu\underline{\mkern-3mu {\bf 3}\mkern-3mu}\mkern3mu}] \rightarrow [2, 0, 1]$.

In the third test case, the array changes as follows during one operation: $[{\mkern3mu\underline{\mkern-3mu {\bf 0}\mkern-3mu}\mkern3mu}, 2] \rightarrow [1, {\mkern3mu\underline{\mkern-3mu {\bf 2}\mkern-3mu}\mkern3mu}] \rightarrow [1, 0]$. And during the second operation: $[{\mkern3mu\underline{\mkern-3mu {\bf 1}\mkern-3mu}\mkern3mu}, 0] \rightarrow [2, {\mkern3mu\underline{\mkern-3mu {\bf 0}\mkern-3mu}\mkern3mu}] \rightarrow [2, 1]$.

 

解题思路

　　看到$k$这么大肯定是有循环节的。在原有数组$a$的情况下令$a_{n+1} = \operatorname{MEX}(a_1, \ldots, a_n)$，此时长度为$n+1$的数组$a$就是$0 \sim n$的一个排列。每轮中的$a_i = \operatorname{MEX}(a_1, \ldots, a_n)$其实就是交换$a_i$和$a_{n+1}$的值，因为在交换后还是$0 \sim n$的一个排列，因此$\operatorname{MEX}(a_1, \ldots, a_n)$还是等于$a_{n+1}$。

　　所以在一轮操作之后，数组从$[a_1, a_2, \ldots, a_{n+1}]$变到$[a_{n+1}, a_1, a_2, \ldots, a_n]$，本质就是循环右移一个单位。所以最终答案就是将原数组补充$a_{n+1} = \operatorname{MEX}(a_1, \ldots, a_n)$后循环右移$k \bmod (n+1)$个单位。

　　AC代码如下：

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

typedef long long LL;

const int N = 1e5 + 10;

int a[N];
bool vis[N];

void solve() {
    int n, m;
    scanf("%d %d", &n, &m);
    memset(vis, 0, n + 10);
    for (int i = 0; i < n; i++) {
        scanf("%d", a + i);
        vis[a[i]] = true;
    }
    for (int i = 0; i <= n; i++) {
        if (!vis[i]) {
            a[n++] = i;
            break;
        }
    }
    m %= n;
    for (int i = 0, j = (n - m) % n; i < n - 1; i++) {
        printf("%d ", a[j]);
        if (++j == n) j = 0;
    }
    printf("\n");
}

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

 

参考资料

　　Pinely Round 2 Editorial：https://codeforces.com/blog/entry/119902