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 For(i,a,b) for (int i=a;i<=b;i++)
using namespace std;
#define int long long

inline int read(){int 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;}
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;

signed main(){
	
	int T=read();
    For(test,1,T){
        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;
        }
        r=1;
        while(1){
            int up=(r+1)/2, dw=r/2;
            int now=r*up+up*dw;
            if(now<n){r++;continue;}
            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;
            }
        }
        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];
            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("");
        }
    }
    
}