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 ll long long
#define scf scanf
using namespace std;
ll t, m, k, a[300000], b[1000][1000], n, sq[10000], sq2[10000], kiri, kanan, tengah, ind;
bool lebih;
pair<ll,ll>p[300000];
int main()
{
	for (ll i = 0; i < 10000; i++)
	{
		if (i % 2 == 0)
		{
			sq[i] = 3 * i * i / 4;
			sq2[i] = i * i / 2;
		}
		else
		{
			sq[i] = i * i - (i - 1) * (i - 1) / 4;
			sq2[i] = i * (i + 1) / 2;
		}
	}
	scf("%lld", &t);
	for (ll i = 0; i < t; i++)
	{
		scf("%lld%lld", &m, &k);
		kiri = 0; kanan = 9999;
		while (kiri <= kanan)
		{
			tengah = (kiri + kanan) / 2;
			if (sq[tengah] < m)
			{
				kiri = tengah + 1;
			}
			else
			{
				kanan = tengah - 1;
			}
		}
		n = kiri;
		lebih = false;
		for (ll j = 1; j <= k; j++)
		{
			scf("%lld", &a[j]);
			p[j - 1] = make_pair(a[j], j);
		}
		sort(p, p+k);
		if (n % 2 == 0)
		{
			if (p[k - 1].first > n * n / 2)
			{
				lebih = true;
			}
		}
		else
		{
			if (p[k - 1].first > n * (n + 1) / 2)
			{
				lebih = true;
			}
		}
		if (lebih)
		{
			kiri = 0; kanan = 9999;
			while (kiri <= kanan)
			{
				tengah = (kiri + kanan) / 2;
				if (sq2[tengah] < p[k - 1].first)
				{
					kiri = tengah + 1;
				}
				else
				{
					kanan = tengah - 1;
				}
			}
			n = kiri;
		}
		for (ll j = 1; j <= n; j++)
		{
			for (ll l = 1; l <= n; l++)
			{
				b[j][l] = 0;
			}
		}
		ind = k - 1;
		for (ll j = 1; j <= n; j+=2)
		{
			for (ll l = 2; l <= n; l+=2)
			{
				if (ind >= 0)
				{
					b[j][l] = p[ind].second;
					p[ind].first--;
				}
				while (ind >= 0)
				{
					if (p[ind].first > 0)
					{
						break;
					}
					else
					{
						ind--;
					}
				}
			}
		}
		for (ll j = 1; j <= n; j+=2)
		{
			for (ll l = 1; l <= n; l+=2)
			{
				if (ind >= 0)
				{
					b[j][l] = p[ind].second;
					p[ind].first--;
				}
				while (ind >= 0)
				{
					if (p[ind].first > 0)
					{
						break;
					}
					else
					{
						ind--;
					}
				}
			}
		}
		for (ll j = 2; j <= n; j+=2)
		{
			for (ll l = 1; l <= n; l+=2)
			{
				if (ind >= 0)
				{
					b[j][l] = p[ind].second;
					p[ind].first--;
				}
				while (ind >= 0)
				{
					if (p[ind].first > 0)
					{
						break;
					}
					else
					{
						ind--;
					}
				}
			}
		}
		printf("%lld\n", n);
		for (ll j = 1; j <= n; j++)
		{
			for (ll l = 1; l <= n; l++)
			{
				printf("%lld", b[j][l]);
				if (l == n)
				{
					printf("\n");
				}
				else
				{
					printf(" ");
				}
			}
		}
	}
}