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;
typedef long long ll;
typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
typedef double db;
#define ls(x) ((x)<<1)
#define rs(x) ((x)<<1|1)
#define low(x) ((x)&-(x))
#define all(x) x.begin(),x.end()
#define mp make_pair
#define X first
#define Y second
#ifdef _DEBUG
    const int N=1e3+10;
#else
    const int N=1e6+10;
#endif
const ll mod=1e9+7;
inline ll gcd(ll a,ll b){return !b?a:gcd(b,a%b);}
inline ll q_pow(ll a,ll x){ll ans=1,tmp=a;while(x){if(x&1)(ans*=tmp)%=mod;(tmp*=tmp)%=mod;x>>=1;}return ans;}
template<typename T> inline void re(T &N){int f=1;char c;while((c=getchar())< '0'||c> '9')if(c=='-')f=-1;N=c-'0';while((c=getchar())>='0'&&c<='9')N=N*10+c-'0';N*=f;}
int m,n,t=1,st,en;
int a[N];
priority_queue<pi>q;
int ans[1010][1010];

int main()
{
    #ifdef _DEBUG
        freopen("data.txt","r",stdin);
    #endif
    re(t);
    while(t--)
    {
        re(m);re(n);
        for(int i=1;i<=n;i++)re(a[i]);
        int key=max_element(a+1,a+n+1)-a;
        for(int i=1;i<=n;i++)if(a[i])q.push({a[i],i});
        int siz=1;
        while(siz*siz-(siz/2)*(siz/2)< m||siz*((siz+1)/2)< a[key])siz++;
        for(int i=1;i<=siz;i++)
            for(int j=1;j<=siz;j++)
                ans[i][j]=0;
        pi la=q.top();q.pop();
        for(int i=2;i<=siz;i+=2)
            for(int j=1;j<=siz;j+=2)
            {
                ans[i][j]=la.Y;
                if(!(--la.X))
                {
                    if(!q.empty())
                        la=q.top(),q.pop();
                    else
                        la={1e9,0};
                }
            }
        for(int i=1;i<=siz;i+=2)
            for(int j=1;j<=siz;j+=2)
            {
                ans[i][j]=la.Y;
                if(!(--la.X))
                {
                    if(!q.empty())
                        la=q.top(),q.pop();
                    else
                        la={1e9,0};
                }
            }
        for(int i=1;i<=siz;i+=2)
            for(int j=2;j<=siz;j+=2)
            {
                ans[i][j]=la.Y;
                if(!(--la.X))
                {
                    if(!q.empty())
                        la=q.top(),q.pop();
                    else
                        la={1e9,0};
                }
            }
        printf("%d\n",siz);
        for(int i=1;i<=siz;i++)
            for(int j=1;j<=siz;j++)
                printf("%d%c",ans[i][j],j==siz?'\n':' ');
    }
}