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;

#define ll long long
#define pb push_back
#define pll pair<ll,ll>
#define ff first
#define ss second
#define endl "\n"
const ll maxn=3e5+10;
const ll base=3e18;
const ll mod=998244353;
const long double eps=1e-10;

pll a[maxn];
ll ans[3003][3003];

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    if (fopen("t.inp", "r"))
    {
        freopen("test.inp", "r", stdin);
        freopen("test.out", "w", stdout) ;
    }
    ll t;
     cin>> t;
     while (t--)
     {
        ll m, k;
         cin>> m>> k;
         for (int i=1;i<=k;i++)
         {
             cin>>a[i].ff;
             a[i].ss=i;
         }
         sort(a+1,a+k+1);
         ll n=0;
         while (n*((n+1)/2)<a[k].ff||n*n-(n/2)*(n/2)<m) n++;
         for (int i=1;i<=n;i++)
         {
             for (int j=1;j<=n;j++)
             {
                 ans[i][j]=0;
             }
         }
         ll pos=k;
         bool kt=false;
         for (int i=1;i<=n;i++)
         {
              if (kt) break;
             for (int j=1;j<=n;j++)
             {
                 if ((i%2==0)&&(j%2==1))
                 {
                     a[pos].ff--;
                     ans[i][j]=a[pos].ss;
                     if (!a[pos].ff&&a[pos-1].ff==0)
                     {
                         kt=true;
                         break;
                     }
                     if (!a[pos].ff) pos--;
                 }
             }

         }
         for (int i=1;i<=n;i++)
         {
             if (kt) break;
             for (int j=1;j<=n;j++)
             {
                 if ((i%2==1)&&(j%2==1))
                 {
                     a[pos].ff--;
                     ans[i][j]=a[pos].ss;
                     if (!a[pos].ff&&a[pos-1].ff==0)
                     {
                         kt=true;
                         break;
                     }
                     if (!a[pos].ff) pos--;
                 }
             }

         }
         for (int i=1;i<=n;i++)
         {
             if (kt) break;
             for (int j=1;j<=n;j++)
             {
                 if ((i%2==1)&&(j%2==0))
                 {
                     a[pos].ff--;
                     ans[i][j]=a[pos].ss;
                     if (!a[pos].ff&&a[pos-1].ff==0)
                     {
                         kt=true;
                         break;
                     }
                     if (!a[pos].ff) pos--;
                 }
             }

         }
         cout <<n<<endl;
         for (int i=1;i<=n;i++)
         {
             for (int j=1;j<=n;j++)
             {
                 cout <<ans[i][j]<<" ";
             }
             cout <<endl;
         }

     }
}