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 INF 1234567890
#define ll long long

int T;
int M, K;
vector<pair<int, int> > v;
int A[640][640];

bool f(int N) // O(N^2) = O(M)
{
	if (N*N < M) return false;
	for(int i=0; i<N; i++)
		for(int j=0; j<N; j++)
			A[i][j] = 0;

	int ty = 0, i = 0, j = 0, di = 0, dj = 0;
	for(auto [cnt,val] : v)
	{
		while(cnt > 0)
		{
			if (ty == 0)
			{
				A[i][j] = val;
				j++;
				if (j >= N) i += 2, j = 0;
				if (i >= N) ty = 1, i = 1, j = 0;
				if (i >= N) ty = 2;
			}
			else if (ty == 1)
			{
				A[i][j] = val;
				// ty=0이랑 ty=1에 걸치는 경우 대각선 처리
				if (i-1>=0 && j-1>=0 && A[i-1][j-1] == val)
				{
					if (di >= N || A[di][dj] == val) return false;
					swap(A[i-1][j-1], A[di][dj]);
					dj += 2; if (dj >= N) di += 2, dj = 0;
				}
				if (i-1>=0 && j+1<N && A[i-1][j+1] == val)
				{
					if (di >= N || A[di][dj] == val) return false;
					swap(A[i-1][j+1], A[di][dj]);
					dj += 2; if (dj >= N) di += 2, dj = 0;
				}
				if (i+1<N && j-1>=0 && A[i+1][j-1] == val)
				{
					if (di >= N || A[di][dj] == val) return false;
					swap(A[i+1][j-1], A[di][dj]);
					dj += 2; if (dj >= N) di += 2, dj = 0;
				}
				if (i+1<N && j+1<N && A[i+1][j+1] == val)
				{
					if (di >= N || A[di][dj] == val) return false;
					swap(A[i+1][j+1], A[di][dj]);
					dj += 2; if (dj >= N) di += 2, dj = 0;
				}

				j += 2;
				if (j >= N) i += 2, j = 0;
				if (i >= N) ty = 2;
			}
			else if (ty == 2) return false;
			cnt--;
		}
	}
	return true;
}

int main()
{
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d %d", &M, &K);
		// clear
		v.clear();

		for(int i=1; i<=K; i++)
		{
			int t;
			scanf("%d", &t);
			if (t) v.push_back({t, i});
		}
		sort(v.rbegin(), v.rend());

		// mid*mid로 가능한가? lo : 불가능, hi : 가능
		int lo = 1, hi = (int)ceil(sqrt(M))*2+1; // 635
		for(int i=0; i<10; i++)
		{
			int mid = lo+hi>>1;
			if (f(mid)) hi = mid;
			else lo = mid;
		}
		printf("%d\n", hi);
		f(hi);
		for(int i=0; i<hi; i++)
		{
			for(int j=0; j<hi; j++)
				printf("%d ", A[i][j]);
			printf("\n");
		}
	}
	return 0;
}