题干：

One day Ms Swan bought an orange in a shop. The orange consisted of n·k segments, numbered with integers from 1 to n·k.

There were k children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the i-th (1 ≤ i ≤ k) child wrote the number ai (1 ≤ ai ≤ n·k). All numbers ai accidentally turned out to be different.

Now the children wonder, how to divide the orange so as to meet these conditions:

• each child gets exactly n orange segments;
• the i-th child gets the segment with number ai for sure;
• no segment goes to two children simultaneously.

Help the children, divide the orange and fulfill the requirements, described above.

Input

The first line contains two integers n, k (1 ≤ n, k ≤ 30). The second line contains kspace-separated integers a1, a2, ..., ak (1 ≤ ai ≤ n·k), where ai is the number of the orange segment that the i-th child would like to get.

It is guaranteed that all numbers ai are distinct.

Output

Print exactly n·k distinct integers. The first n integers represent the indexes of the segments the first child will get, the second n integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces.

You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them.

Examples

Input

2 2
4 1

Output

2 4 
1 3 

Input

3 1
2

Output

3 2 1 

题目大意：

一个橘子分成n*k块（对应编号1~n*k），k个人分，每个人都有一块最想得到的编号，现在要你每个人都分到n块，应该怎么分。

解题报告：

特殊块特殊给，剩下的平均分。直接用整除，，算是个小技巧了吧、、反正数据量也小，，怎么搞都行。。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int n,k; 
int a[MAX];
int opa[MAX];
int main()
{cin>>n>>k;memset(opa,-1,sizeof opa);for(int i = 1; i<=k; i++) {scanf("%d",a+i);opa[a[i]] = i;}if(n == 1) {for(int i = 1; i<=k; i++) {printf("%d\n",a[i]);}return 0 ;}int cur = 0;for(int i = 1; i<=n*k; i++) {if(opa[i] != -1) continue;opa[i] = cur/(n-1)+1;cur++;}int times = 0;while(1) {times++;if(times == k+1) break;for(int i = 1; i<=n*k; i++) {if(opa[i] == times) {printf("%d ",i);}}printf("\n");}return 0 ;}