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:
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#include <bits/stdc++.h>

#define mp make_pair
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()

using namespace std;

typedef long long LL;
typedef pair<int,int> PII;
typedef vector < int > VI;
typedef double D;

const int MN = 100005, MX = 5005, inf = 1000000005, mod = 1000000007;
const LL INF = 1000000000000000005LL;

vector < PII > wek, wek2;

int ans[MX][MX];

int dx[] = {0, 0, 1, 1};
int dy[] = {0, 1, 0, 1};

int wez(bool keep = false)
{
	if(!wek2.size())
		return 0;
	while(wek2.back().fi == 0)
		wek2.pop_back();
	if(!wek2.size())
		return 0;
	int ret = wek2.back().se;
	if(!keep)
		wek2.back().fi--;
	return ret;
}

bool check(int N, bool write = false)
{
	wek2.clear();
	for(auto w : wek)
		wek2.pb(w);
	vector < PII > gdzie;
	for(int i = 2; i <= N; i += 2)
	{
		for(int j = 1; j <= N; j += 2)
		{
			int cur = wez();
			if(!cur)
				break;
			ans[i][j] = cur;
			gdzie.pb({i, j});
		}
		if(!wez(true))
			break;
	}
	for(int i = 1; i <= N; i += 2)
	{
		for(int j = 1; j <= N; j += 2)
		{
			int cur = wez();
			if(!cur)
				break;
			ans[i][j] = cur;
			gdzie.pb({i, j});
		}
		if(!wez(true))
			break;
	}
	for(int i = 1; i <= N; i += 2)
	{
		for(int j = 2; j <= N; j += 2)
		{
			int cur = wez();
			if(!cur)
				break;
			ans[i][j] = cur;
			gdzie.pb({i, j});
		}
		if(!wez(true))
			break;
	}
	bool ret = true;
	if(wez(true))
		ret = false;
	for(auto g : gdzie)
		for(int i = 0; i < 4; ++i)
		{
			PII to_check = {g.fi + dx[i], g.se + dy[i]};
			if(min(to_check.fi, to_check.se) > 1 && max(to_check.fi, to_check.se) <= N)
			{
				bool ok = ((ans[to_check.fi - 1][to_check.se - 1] == 0 || ans[to_check.fi - 1][to_check.se - 1] != ans[to_check.fi][to_check.se]) && (ans[to_check.fi - 1][to_check.se] == 0 || ans[to_check.fi - 1][to_check.se] != ans[to_check.fi][to_check.se - 1]));
				//printf("%d %d %d\n", to_check.fi, to_check.se, (int)ok);
				if(!ok)
					ret = false;
			}
		}
	if(!write)
		for(auto g : gdzie)
			ans[g.fi][g.se] = 0;
	return ret;
		
}

int bs()
{
	int zd = 1, zg = MX - 3;
	while(zd != zg)
	{
		int mid = (zd + zg) >> 1;
		if(check(mid))
			zg = mid;
		else
			zd = mid + 1;
	}
	return zd;
}

void solve()
{
	int m, k;
	scanf("%d%d", &m, &k);
	for(int i = 1; i <= k; ++i)
	{
		int ile;
		scanf("%d", &ile);
		wek.pb({ile, i});
	}
	sort(all(wek));
	int n = bs();
	check(n, true);
	printf("%d\n", n);
	for(int i = 1; i <= n; ++i)
	{
		for(int j = 1; j <= n; ++j)
		{
			printf("%d ", ans[i][j]);
			ans[i][j] = 0;
		}
		printf("\n");
	}
	wek.clear();
}

int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
		solve();
}