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:
/*input

*/

#include <iostream>
#include <vector>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cassert>
#include <algorithm>
#include <cstdlib>
#include <numeric>
#include <utility>
#include <tuple>
#include <climits>
#include <fstream>
#include <bitset>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <stack>
#include <queue>
#include <random>
#include <chrono>
#include <ios>
#include <iomanip>
#include <functional>
#include <array>

using namespace std;

#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define FORA(i, a) for (auto &i : a)
#define FORB(i, a, b) for (int i = a; i >= b; --i)
#define SZ(a) ((int) a.size())
#define ALL(a) begin(a), end(a)

typedef int64_t ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
#define fi first
#define se second

// start of code

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

	int t;
	cin >> t;
	while (t--) {
		int m, k;
		cin >> m >> k;
		vector<int> a(k);
		FORA(x, a) cin >> x;
		int mx = *max_element(ALL(a)), n = 1;
		for (; m > n * n - (n / 2) * (n / 2) || mx > n * ((n + 1) / 2); ++n);
		vector<pair<int, int>> vec;
		for (int i = 1; i < n; i += 2) for (int j = 0; j < n; j += 2) vec.emplace_back(i, j);
		for (int i = 0; i < n; i += 2) for (int j = 0; j < n; j += 2) vec.emplace_back(i, j);
		for (int i = 0; i < n; i += 2) for (int j = 1; j < n; j += 2) vec.emplace_back(i, j);
		vector<vector<int>> b(n, vector<int>(n, 0));
		vector<int> c(k);
		iota(ALL(c), 0);
		sort(ALL(c), [&](int i, int j) { return a[i] > a[j]; });
		FORA(i, c) {
			while (a[i]--) {
				b[vec.back().first][vec.back().second] = i + 1;
				vec.pop_back();
			}
		}
		cout << n << '\n';
		FORA(v, b) {
			FORA(x, v) {
				cout << x << ' ';
			}
			cout << '\n';
		}
	}

	return 0;
}