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>
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
#define pb push_back
#define ALL(s) s.begin(),s.end()
#define FIN ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define SZ(s) int(s.size())
using namespace std;
typedef long long ll;
typedef pair<int,int> ii;

void bad(){
	cout<<"-1\n";
	exit(0);
}


const int MAXN=1010;
int ans[MAXN][MAXN],n;

bool can(int i, int j, int x){
	if(ans[i][j])return 0;
	if(i&&j&&ans[i-1][j-1]==x)return 0;
	if(i&&j+1<n&&ans[i-1][j+1]==x)return 0;
	if(i+1<n&&j&&ans[i+1][j-1]==x)return 0;
	if(i+1<n&&j+1<n&&ans[i+1][j+1]==x)return 0;
	return 1;
}

int main(){FIN;
	int t; cin>>t;
	while(t--){
		int m,k,mx=0; cin>>m>>k;
		vector<ii> v(k);
		fore(i,0,k) cin>>v[i].fst,mx=max(mx,v[i].fst),v[i].snd=i+1;
		
		n=1;
		while(1){
			int up=(n+1)/2, dw=n/2;
			int now=n*up+up*dw;
			
			//no entran todas
			if(now<m){n++;continue;}
			
			//no entran las del maximo
			int has=(mx+n-1)/n;
			if(2*has-1>n){n++;continue;}
			break;
			
		}
		
		sort(ALL(v)); reverse(ALL(v));
		
		vector<int> a;
		for(auto x:v) fore(i,0,x.fst) a.pb(x.snd);
		reverse(ALL(a));
		
		
		
		
		//clean
		fore(i,0,n) fore(j,0,n) ans[i][j]=0;
		
		//poner maximo
		for(int i=0;i<n;i+=2){
			fore(j,0,n) if(j%2==1&&SZ(a)) ans[i][j]=a.back(),a.pop_back();
		}
		for(int i=0;i<n;i+=2){
			fore(j,0,n) if(j%2==0&&SZ(a)) ans[i][j]=a.back(),a.pop_back();
		}
		
		//completar
		fore(i,0,n) fore(j,0,n){
			if(i%2==1&&j%2==1)continue;
			if(SZ(a)&&can(i,j,a.back())) ans[i][j]=a.back(),a.pop_back();
		}
		
		
		for(int i=1;i<n;i+=2){
			fore(j,0,n) if(j%2==0&&SZ(a)) ans[i][j]=a.back(),a.pop_back();
		}
		
		int ck=1;
		//check
		if(ck){
			if(SZ(a)) bad();
			fore(i,0,n-1) fore(j,0,n-1){
				int am=(ans[i][j]==0)+(ans[i+1][j]==0)+(ans[i][j+1]==0)+(ans[i+1][j+1]==0);
				if(!am)bad();
				if(ans[i][j]&&ans[i+1][j+1]) if(ans[i][j]==ans[i+1][j+1]) bad();
				if(ans[i+1][j]&&ans[i][j+1]) if(ans[i+1][j]==ans[i][j+1]) bad();
			}
		}
		
		cout<<n<<"\n";
		fore(i,0,n) fore(j,0,n) cout<<ans[i][j]<<" \n"[j+1==n];
	}
}