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>
#pragma GCC optimize(2)
using namespace std;
#define ll long long
#define int long long
inline ll read(){ll x=0,f=1;char ch=getchar();for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-f;for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';return x*f;}
inline void write(ll x){if(x<0)x=-x,putchar('-');if(x>9)write(x/10);putchar(x%10+'0');}
inline void writeln(ll x){write(x);puts("");}
const int N=100005;
struct point{
	int x,i;
}a[N];
int ans[1005][1005];
bool cmp(point a,point b)
{
	return a.x>b.x;
}
int ox,oy;
int ex,ey;
int r;
void fill_in_odd(int x,int t)
{
	while(t>0){
		if(ans[ox][oy]==0)ans[ox][oy]=x,t--;
		ox++;
		if(ox>r){
			oy=oy+2;
			ox=1;
		}
	}
}
void fill_in_even(int x,int t)
{
	while(t>0){
		ans[ex][ey]=x;
		t--;
		ex-=2;
		if(ex<=0){
			ex=r;
			ey-=2;
		}
		if(ey<=0){
			if(r%2==0){
				ex=r,ey=r-1;
			}else {
				ex=r,ey=r;
			}
		}
	}
}
int b[N],c[N];
int lb,lc;
void solve()
{
	int n,m;
	n=read();m=read();
	for(int i=1;i<=m;i++){
		a[i].x=read();
		a[i].i=i;
	}
	sort(a+1,a+m+1,cmp);
	int now=a[1].x;
	int tmp=n*2/3;
	int las=-1;
	lb=0,lc=0;
	b[++lb]=1;
	for(int i=2;i<=m;i++){
		if(now+a[i].x>=tmp)c[++lc]=las,las=i;
		else now=now+a[i].x,b[++lb]=i;
	}
//	int s=max(now,(n/3+(n%3!=0))*2);
//	r=sqrt(s);
//	while(r*(r/2+(r%2!=0))<s)r++;
		r=1;
		while(1){
			int up=(r+1)/2, dw=r/2;
			int now=r*up+up*dw;
			
			//no entran todas
			if(now<n){r++;continue;}
			
			//no entran las del maximo
			int has=(a[1].x+r-1)/r;
			if(2*has-1>r){r++;continue;}
			break;
			
		}
	if(n==1)r=1;
	ox=1;oy=1;
	if(r%2==0){
		ex=r,ey=r;
	}else {
		ex=r,ey=r-1;
	}
	for(int i=1;i<=r;i++){
		for(int j=1;j<=r;j++){
			ans[i][j]=0;
		}
	}
//	cout<<' '<<a[las].i<<' '<<tmp<<' '<<now<<endl;
	fill_in_even(a[las].i,a[las].x);
	for(int i=2;i<=lc;i++){
		int k=c[i];
		fill_in_even(a[k].i,a[k].x);
	}
	for(int i=1;i<=lb;i++){
		int k=b[i];
//		cout<<a[k].i<<' '<<a[k].x<<endl;
		fill_in_odd(a[k].i,a[k].x);
	}
	printf("%d\n",r);
	for(int i=1;i<=r;i++){
		for(int j=1;j<=r;j++){
			printf("%d ",ans[i][j]);
		}
		puts("");
	}
}
signed main()
{
	int T=read();
	while(T--){
		solve();
	}
}

/*
3
28 3
5 13 10
29 3
8 12 9
25 3
12 10 3
*/