Problem: You like numbers, don't you? Nastia has a lot of numbers and she wants to share them with you! Isn't it amazing?

Let a_i be how many numbers i (1 ≤ i ≤ k) you have.

An n × n matrix is called beautiful if it contains all the numbers you have, and for each 2 × 2 submatrix of the original matrix is satisfied: 

  1. The number of occupied cells doesn't exceed 3; 
  2. The numbers on each diagonal are distinct. 



Make a beautiful matrix of minimum size.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains 2 integers m and k (1 ≤ m, k ≤ 10^5) — how many numbers Nastia gave you and the length of the array a, respectively.

The second line of each test case contains k integers a_1, a_2, …, a_{k} (0 ≤ a_i ≤ m, a_1 + a_2 + … + a_{k} = m), where a_i is how many numbers i you have.

It's guaranteed that the sum of m and k in one test doesn't exceed 2 ⋅ 10^5.

Output

For each t test case print a single integer n — the size of the beautiful matrix.

In the next n lines print n integers b_{i, j} (0 ≤ b_{i, j} ≤ k; if position is empty, print b_{i, j} = 0) — the beautiful matrix b you made up. 

Example

Input


2
3 4
2 0 0 1
15 4
2 4 8 1


Output


2
4 1
0 1
5
3 0 0 2 2
3 2 3 3 0
0 1 0 4 0
3 0 0 0 0
2 1 3 3 3

Note

Note that 0 in this problem represents a blank, not a number.

Examples of possible answers for the first test case:

\begin{array}{cc} 1 & 1 \\\ 4 & 0 \\\ \end{array} \hspace{0,5cm} \begin{array}{cc} 1 & 4 \\\ 1 & 0 \\\ \end{array} \hspace{0,5cm} \begin{array}{cc} 4 & 0 \\\ 1 & 1 \\\ \end{array}

Examples of not beautiful matrices for the first test case:

\begin{array}{cc} 1 & 0 \\\ 4 & 1 \\\ \end{array} \hspace{0,5cm} \begin{array}{cc} 4 & 1 \\\ 7 & 1 \\\ \end{array} \hspace{0,5cm} \begin{array}{cc} 1 & 0 \\\ 4 & 0 \\\ \end{array}

The example of the not beautiful matrix for the second test case:

\begin{array}{cc} 3 & 4 & 0 & 2 & 2 \\\ 3 & 2 & 3 & 3 & 0 \\\ 0 & 1 & 0 & 0 & 0 \\\ 3 & 0 & 0 & 0 & 0 \\\ 2 & 1 & 3 & 3 & 3 \\\ \end{array}

Everything is okay, except the left-top submatrix contains 4 numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
# define rep(i,a,b) for(int i=(a); i<=(b); ++i)
# define drep(i,a,b) for(int i=(a); i>=(b); --i)
typedef long long int_;
inline int readint(){
	int a = 0; char c = getchar(), f = 1;
	for(; c<'0'||c>'9'; c=getchar())
		if(c == '-') f = -f;
	for(; '0'<=c&&c<='9'; c=getchar())
		a = (a<<3)+(a<<1)+(c^48);
	return a*f;
}

const int MaxN = 100005;
const int SqrtN = 500;

struct Node{
	int cnt, val;
	bool operator < (const Node &t) const {
		return cnt > t.cnt;
	}
};
Node num[MaxN];
int maze[SqrtN][SqrtN];

int main(){
	for(int T=readint(); T; --T){
		int m = readint(), k = readint();
		rep(i,1,k){
			num[i].cnt = readint();
			num[i].val = i;
		}
		sort(num+1,num+k+1); int n = 1;
		while(n*n-(n>>1)*(n>>1) < m) ++ n;
		while((n>>1)*((n+1)>>1) // red
			+(n*n+1)/2 // black
			-(n>>1)*(n>>1) < num[1].cnt)
				++ n; // increase
		/* step one: fuck one num */ ;
		for(int i=2; i<=n; i+=2)
		for(int j=1; j<=n&&num[1].cnt; j+=2)
			maze[i][j] = num[1].val, -- num[1].cnt;
		for(int i=1; i<=n; i+=2)
		for(int j=1; j<=n&&num[1].cnt; j+=2)
			maze[i][j] = num[1].val, -- num[1].cnt;
		/* step two: fill in all num */ ;
		int p = 2; // which num is dealt with
		for(int i=2; i<=n; i+=2)
		for(int j=1; j<=n&&p<=k; j+=2){
			if(maze[i][j]) continue;
			for(; !num[p].cnt; ++p)
				if(p == k+1) break;
			if(p != k+1){
				maze[i][j] = num[p].val;
				-- num[p].cnt;
			}
		}
		for(int i=1; i<=n; i+=2)
		for(int j=1; j<=n&&p<=k; j+=2){
			if(maze[i][j]) continue;
			for(; !num[p].cnt; ++p)
				if(p == k+1) break;
			if(p != k+1){
				maze[i][j] = num[p].val;
				-- num[p].cnt;
			}
		}
		for(int i=1; i<=n; i+=2)
		for(int j=2; j<=n&&p<=k; j+=2){
			for(; !num[p].cnt; ++p)
				if(p == k+1) break;
			if(p != k+1){
				maze[i][j] = num[p].val;
				-- num[p].cnt;
			}
		}
		printf("%d\n",n);
		/* step three: output */ ;
		for(int i=1; i<=n; ++i,puts(""))
		for(int j=1; j<=n; ++j){
			printf("%d ",maze[i][j]);
			maze[i][j] = 0; // clear
		}
	}

	return 0;
}

/*

black = (n*n+1)>>1

blank = (n>>1)*(n>>1)

free = black - blank

red = (n>>1)*((n+1)>>1)

*/