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;
 
typedef long long ll;
#define pb push_back
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);   
    
    int t;
    cin >> t;
    
    while(t--){
        vector<pair<int, int>> amts;
        int m,k;
        cin >> m >> k;
        int mx = 0;
        for (int i = 0; i < k; i++){
            int a;
            cin >> a;
            amts.pb(make_pair(-a,i+1));
            if (a > mx) mx = a;
        }
        sort(amts.begin(), amts.end());
        
        int lo = 0;
        int hi = 1023;
        while (hi > lo){
            int n = lo + (hi - lo)/2;
            
            if (m <= n * n - (n/2)*(n/2) && mx <= n * (n - n/2)){
                hi = n;
            }
            else{
                lo = n+1;
            }
        }
        
        int n = lo;
        cout << n << '\n';
        
        vector<vector<int>> board;
        for(int i = 0; i < n; i++){
            vector<int> nw (n, 0);
            board.pb(nw);
        }
        
        int c1 = (n / 2) * (n - n/2);
        int c2 = (n / 2) * (n - n/2);
        int c3 = (n - n / 2) * (n - n/2);
        
        assert (c1 + c2 + c3 + (n/2)*(n/2) == n * n);
        
        for(int i = 0; i < k; i++){
            auto p = amts[i];
            int ct = -p.first;
            int ind = p.second;
            bool br = false;
            for(int j = 0; j < ct; j ++){
                if (c1){
                    c1 --;
                    int x = c1 / (n/2);
                    int y = c1 % (n/2);
                    board[2*x][2*y+1] = ind;
                    if (c1 == 0) br = true;
                }
                else if (br || c2 == 0){
                    c3 --;
                    int x = c3 / (n - n/2);
                    int y = c3 % (n - n/2);
                    board[2*x][2*y] = ind;
                }
                else {
                    c2 --;
                    int x = c2 / (n/2);
                    int y = c2 % (n/2);
                    board[2*y+1][2*x] = ind;
                }
            }
        }
        
        
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                cout << board[i][j] << ' ';
            }
            cout << '\n';
        }
    }
}