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 <iostream>
#include <vector>
#include <cassert>
#include <algorithm>
#include <utility>
using namespace std;
#define dups first
#define num second
#define row first
#define col second

using pi = pair<int,int>;

int m,k;
vector<pi> a;
vector<pi> x,y,z;
vector<vector<int>> ans;

void fastIO() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); }

void takeInput() {
	cin >> m >> k;
	a.resize(k);
	// a[i] = {# dups of actual number, actual number}
	for (int i=0; i<k; i++) {
		cin >> a[i].dups, a[i].num = i+1;
	} sort(a.begin(), a.end(), greater<pair<int,int>>());
}

void constructGroups(int n) {
	for (int i=0; i<n; i++)
	for (int j=0; j<n; j++) {
		if ((i+j) % 2 == 1) {
			if (i % 2 == 0) x.push_back({i,j});
			else y.push_back({i,j});
		} else {
			if (i%2 == 0) z.push_back({i,j});
		}
	}
}

void constructBeautifulMatrix(int n) {
	ans.resize(n);
	for (int i=0; i<n; i++) {
		ans[i].resize(n);
		for (int j=0; j<n; j++) ans[i][j] = 0;
	}

	for (int i=0; i<k; i++) {
		vector<pi>& cur = ((x.empty()) ? y : x);

		while (a[i].dups && !cur.empty()) {
			pi pos = cur.back();
			ans[pos.row][pos.col] = a[i].num;
			cur.pop_back(), a[i].dups--;
		}

		while (a[i].dups--) {
			assert((int)z.size() > 0); /* if z.size() 0 abort, infeasible */
			pi pos = z.back();
			ans[pos.row][pos.col] = a[i].num;
			z.pop_back();
		}
	}
}

void printAnswer(int n) {
	cout << n << '\n';
	for (int i=0; i<n; i++) {
		for (int j=0; j<n; j++) {
			cout << ans[i][j] << ' ';
		} cout << '\n';
	}
}

void solution() {
	int mx = a[0].dups;
	for (int n=1; n<=m; n++) {
		if (mx > n * ((n+1) / 2)) continue;
		if (m > n*n - (n/2)*(n/2)) continue;
		x.clear(), y.clear(), z.clear();
		constructGroups(n);
		constructBeautifulMatrix(n);
		printAnswer(n);
		break;
	}
}

int main() {
	fastIO();
	int t; cin >> t;
	while (t--) {
		takeInput();
		solution();
	}
}