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 ;
#define input freopen( "test.in" , "r" , stdin ) ;
#define stream ios::sync_with_stdio(false) ; cin.tie(0) ; cout.tie(0) ;

const int maxn = 1e5+5 ;

struct node
{
    int num ;
    int p ;
    bool operator < ( const node a ) const {
        return a.num > num ;
    }
}num[maxn];
bool cmp ( node a , node b )
{
    return a.num > b.num ;
}
int main ()
{
    stream
    ll t , sum , n , mx , sn , cnt ,c ;
    cin >> t ;
    while ( t-- ) {
        cin >> sum >> n;
        for (int i = 1; i <= n; i++) {
            cin >> num[i].num;
            num[i].p = i;
        }
        sort(num + 1, num + n + 1 , cmp ) ;
        sn = 1;
        mx = (*max_element( num + 1 , num + n + 1 )).num ;
        while ( sum > sn * sn - (sn / 2) * (sn / 2) || mx > sn * (sn - ( sn / 2 ) ) ) {
            sn++ ;
        }
        cout << sn << endl;
        int MAP[sn+1][sn+1] ;
        c = 1 ;
        for ( int i = 2 ; i <= sn ; i+=2 )
        {
            for ( int j = 1 ; j <= sn ; j++ )
            {
                if ( i % 2 == 0 && j % 2 == 0 )
                {
                    MAP[i][j] = 0 ;
                }
                else
                {
                    while ( num[c].num == 0 && c <= n )
                    {
                        c++ ;
                    }
                    if ( c == n + 1 )
                    {
                        MAP[i][j] = 0 ;
                    }
                    else
                    {
                        MAP[i][j] = num[c].p ;
                        num[c].num-- ;
                    }
                }
            }
        }
        for ( int i = 1 ; i <= sn ; i += 2 )
        {
            for ( int j = 1 ; j <= sn ; j+=2 )
            {
                while ( num[c].num == 0 && c <= n )
                {
                    c++ ;
                }
                if ( c == n + 1 )
                {
                    MAP[i][j] = 0 ;
                }
                else
                {
                    MAP[i][j] = num[c].p ;
                    num[c].num-- ;
                }
            }
        }
        for ( int i = 1 ; i <= sn ; i += 2 )
        {
            for ( int j = 2 ; j <= sn ; j+=2 )
            {
                while ( num[c].num == 0 && c <= n )
                {
                    c++ ;
                }
                if ( c == n + 1 )
                {
                    MAP[i][j] = 0 ;
                }
                else
                {
                    MAP[i][j] = num[c].p ;
                    num[c].num-- ;
                }
            }
        }
        for ( int i = 1 ; i <= sn ; i++ )
        {
            for ( int j = 1 ; j <= sn ; j++ )
            {
                if ( j == 1 )
                {
                    cout << MAP[i][j] ;
                }
                else
                {
                    cout << " " << MAP[i][j] ;
                }
            }
            cout << endl ;
        }
    }
    return 0 ;
}