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>
#define lsb(x) (x & (-x))
#define ll long long
#define ull unsigned long long
#define Test(tt) cout << "Case #" << tt << ": "
 
 
using namespace std;

constexpr int MAXN = 2000;

int mat[MAXN + 1][MAXN + 1];


int main() {
#ifdef HOME
    ifstream cin("A.in");
    ofstream cout("A.out");
#endif
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int t;
    cin >> t;
    while (t--) {
        int m, k;
        cin >> m >> k;
        vector<pair<int, int>> freq(k);
        for (int i = 0; i < k; i++) {
            cin >> freq[i].first;
            freq[i].second = i + 1;
        }
        sort(freq.rbegin(), freq.rend());

        auto check = [&](int n) -> bool {
            int blue = 0, red = 0, yellow = 0;
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    mat[i][j] = 0;
                    if (i % 2 == 1 && j % 2 == 1) {
                        blue++;
                    } else if (i % 2 == 0 && j % 2 == 1) {
                        red++;
                    } else if (i % 2 == 1 && j % 2 == 0) {
                        yellow++;
                    }
                }
            }

            if (blue + red + yellow < m || red + blue < freq[0].first) {
                return false;
            }

            int pos = 0, cur = 0;
            for (auto it : vector<pair<int, int>>{{0, 1}, {1, 1}, {1, 0}}) {
                int x = it.first;
                int y = it.second;
                for (int i = 2 - x; i <= n && pos < k; i += 2) {
                    for (int j = 2 - y; j <= n && pos < k; j += 2) {
                        if (cur < freq[pos].first) {
                            cur++;
                            mat[i][j] = freq[pos].second;
                        } else {
                            pos++;
                            while (pos < k && freq[pos].first == 0) {
                                pos++;
                            }
                            if (pos < k) {
                                cur = 1;
                                mat[i][j] = freq[pos].second;
                            }
                        }
                    }
                }
            }

            return true;
        };

        int n = 0;
        for (int step = 1 << 9; step; step >>= 1) {
            if ((n + step) * (n + step) <= 4 * m && check(n + step) == 0) {
                n += step;
            }
        }

        cout << ++n << "\n";
        
        check(n);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                cout << mat[i][j] << " ";
            }
            cout << "\n";
        }

        // check(2);
    }

    return 0;
}