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;
typedef pair<int, int> pii;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const int N = (int)1e5 + 10;

int A[N];
const int M = 2005;
int F[M][M];

void solve(){
    int m, k;
    cin >> m >> k;
    int mx = 0;
    for(int i = 1; i <= k ; i ++ ){
        cin >> A[i];
        mx = max(mx, A[i]);
    }
    for(int n = 1; n <= m ; n ++ ){
        if(mx <= n * ((n+1)/2) && m <= n * n - (n/2)*(n/2)){
            vector<pii> ord;
            for(int i = 1; i <= k ;i ++ ){
                ord.push_back(mp(A[i], i));
            }
            sort(ord.begin(), ord.end());
            reverse(ord.begin(), ord.end());
            vector<pii> cells;
            for(int i = 1; i <= n; i += 2){
                for(int j = 2; j <= n; j += 2){
                    cells.push_back(mp(i, j));
                }
            }
            for(int i = 1; i <= n; i += 2){
                for(int j = 1; j <= n; j += 2){
                    cells.push_back(mp(i, j));
                }
            }
            for(int i = 2; i <= n; i += 2){
                for(int j = 1; j <= n; j += 2){
                    cells.push_back(mp(i, j));
                }
            }
            for(int i = 1; i <= n; i ++ ){
                for(int j = 1; j <= n; j ++ ){
                    F[i][j]=0;
                }
            }
            int cur = 0;
            for(auto x : ord){
                for(int j = 0 ; j < x.fi; j ++ ){
                    F[cells[cur].fi][cells[cur].se] = x.se;
                    cur ++ ;
                }
            }
            cout << n << "\n";
            for(int i = 1; i <= n; i ++ ){
                for(int j = 1; j <= n; j ++){
                    cout << F[i][j] << " ";
                }
                cout << "\n";
            }
            return;
        }
    }
}

int main(){
    fastIO;
    int tc;
    cin >> tc;
    for(int iq = 1; iq <= tc; iq ++ ){
        solve();
    }
    return 0;
}