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 oo 1000000010
const int N = 200010;
const int M = 1010;
int m , k , arr[N] , mx;

int grid[M][M];

bool check(int n){
	int all = 0 , a = 0 , b = 0 , c = 0;
	for(int i = 0 ;i < n;i++){
		for(int j = 0 ;j < n;j++){
			if((i & 1) && (j & 1)) continue;
			all++;
			if((i + j) & 1){
				if(i & 1)
					a++;
				else
					b++;
			}
			else
				c++;	
			if(all >= m && c + max(a , b) >= mx)
				return true;
		}
	}
	return false;
}

void get(int n){
	//cout << "get " << n << endl;
	vector< pair< int , int > > v1 , v2 , v3;
	vector< pair< int , int > > v;
	for(int i = 0 ;i < n;i++){
		for(int j = 0 ;j < n;j++){
			grid[i][j] = 0;
			if((i & 1) && (j & 1)) continue;
			if((i + j) & 1){
				if(i & 1)
					v1.push_back(make_pair(i , j));
				else
					v2.push_back(make_pair(i , j));
			}
			else
				v3.push_back(make_pair(i , j));
		}
	}
	//cout << (int)v1.size() + (int)v2.size() + (int)v3.size() << endl;
	if((int)v1.size() < (int)v2.size()) swap(v1 , v2);
	for(int i = 1;i <= k;i++){
		v.push_back(make_pair(arr[i] , i));
	}
	sort(v.begin(),v.end());
	for(int r , c , i = (int)v.size() - 1 ;i >= 0;i--){
		for(int j = 0 ;j < (int)v[i].first;j++){
			if((int)v1.size() > 0){
				r = v1.back().first;
				c = v1.back().second;
				grid[r][c] = v[i].second;
				v1.pop_back();
			}
			else if((int)v3.size() > 0){
				r = v3.back().first;
				c = v3.back().second;
				grid[r][c] = v[i].second;
				v3.pop_back();
			}
			else{
				assert((int)v2.size() > 0);
				r = v2.back().first;
				c = v2.back().second;
				grid[r][c] = v[i].second;
				v2.pop_back();
			}
		}
	}
}

void solve(){
	scanf("%d%d",&m,&k);
	mx = 0;
	for(int i = 1;i <= k;i++){
		scanf("%d",&arr[i]);
		mx = max(mx , arr[i]);
	}
	int low = 1 , high = M - 1 , res = M - 1 , mid;
	while(high >= low){
		mid = (low + high) / 2;
		if(check(mid))
			res = mid , high = mid - 1;
		else
			low = mid + 1;
	}
	printf("%d\n",res);
	get(res);
	for(int i = 0 ;i < res;i++){
		for(int j = 0 ;j < res;j++){
			if(j) putchar(' ');
			printf("%d",grid[i][j]);
		}
		puts("");
	}
}



int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		solve();
	}
	return 0;
}