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 N 100010
typedef long long ll;
inline ll read(){
	ll x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int T,m,k,a[N],c[1234][1234],p[N];
inline bool check(int n){
	return n*n-(n>>1)*(n>>1)>=m && n*((n+1)>>1)>=a[p[1]];	
}
void Solve(){
	m=read(),k=read();
	for(int i=1;i<=k;++i){
		a[i]=read();
		p[i]=i;
	}
	sort(p+1,p+k+1,[&](int i,int j){return a[i]>a[j];});
	int n=1;
	while(!check(n))++n;
	for(int i=1;i<=n;++i){
		for(int j=1;j<=n;++j){
			c[i][j]=0;
		}
	}
	static vector<pair<int,int> > pos;
	pos.clear();
	for(int i=1;i<=n;i+=2){
		for(int j=2;j<=n;j+=2){
			pos.emplace_back(i,j);
		}
	}
	for(int i=1;i<=n;i+=2){
		for(int j=1;j<=n;j+=2){
			pos.emplace_back(i,j);
		}
	}
	for(int i=2;i<=n;i+=2){
		for(int j=1;j<=n;j+=2){
			pos.emplace_back(i,j);
		}
	}
	reverse(pos.begin(),pos.end()); 
	for(int t=1;t<=k;++t){
		int x=p[t];
		while(a[x]--){
			c[pos.back().first][pos.back().second]=x;
			pos.pop_back();
		}
	}
	printf("%d\n",n);
	for(int i=1;i<=n;++i){
		for(int j=1;j<=n;++j){
			printf("%d ",c[i][j]);
		}
		printf("\n");
	}
}
int main(){
	T=read();
	while(T--){
		Solve();
	}
	return 0;
}