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<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#include<assert.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define pb push_back
#define fi first
#define se second
const int N=1e5+10,M=1e3+10;
int t,m,k,a[N],b[M][M],mx,x[N],sum;
vector<P>odd,even,spe;
bool ok(int x){
    int y=(x+1)/2;
    return 1ll*x*y>=mx && 1ll*x*y+1ll*y*(x-y)>=sum;
}
bool cmp(int x,int y){
    return a[x]>a[y];
}
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&m,&k);
        mx=0;sum=0;
        for(int i=1;i<=k;++i){
            scanf("%d",&a[i]);
            sum+=a[i];
            x[i]=i;
            mx=max(mx,a[i]);
        }
        sort(x+1,x+k+1,cmp);
        int l=1,r=1e6;
        while(l<=r){
            int mid=(l+r)/2;
            if(ok(mid)){
                r=mid-1;
            }
            else{
                l=mid+1;
            }
        }
        printf("%d\n",l);
        int num=0;
        odd.clear();even.clear();spe.clear();
        for(int i=1;i<=l;++i){
            for(int j=1;j<=l;++j){
                if(i%2==0 && j%2==0){
                    b[i][j]=0;
                }
                else{
                    if(i&1){
                        if(mx<=1/2*num){
                            odd.push_back(P(i,j));
                        }
                        else{
                            if(j%2==0)spe.push_back(P(i,j));
                            else odd.push_back(P(i,j));
                        }
                    }
                    else{
                        even.push_back(P(i,j));
                    }
                    b[i][j]=0;
                    num++;
                }
            }
        }
        int now=1;
        if(mx<=1/2*num){
            while(now<=k){
                while(now<=k && !a[x[now]])now++;
                if(now>k)break;
                int v=x[now];
                a[v]--;
                if(odd.size()){
                    P z=odd.back();odd.pop_back();
                    b[z.fi][z.se]=v;
                }
                else{
                    P z=even.back();even.pop_back();
                    b[z.fi][z.se]=v;
                }
            }
        }
        else{
            while(now<=k){
                while(now<=k && !a[x[now]])now++;
                if(now>k)break;
                int v=x[now];
                a[v]--;
                if(spe.size()){
                    P z=spe.back();spe.pop_back();
                    b[z.fi][z.se]=v;
                }
                else if(odd.size()){
                    P z=odd.back();odd.pop_back();
                    b[z.fi][z.se]=v;
                }
                else{
                    P z=even.back();even.pop_back();
                    b[z.fi][z.se]=v;
                }
            }
        }
        for(int i=1;i<=l;++i){
            for(int j=1;j<=l;++j){
                printf("%d%c",b[i][j]," \n"[j==l]);
            }
        }
    }
    return 0;
}
/*
2
1
00
11
01
3
011001
111010
010001
*/