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:
#ifndef _MY
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
#endif
//#define _GLIBCXX_DEBUG
#include "bits/stdc++.h"
#define F first
#define S second
#define mp make_pair
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef long double ld;
const int64_t INF = (int64_t)(2e18);
const int inf = (int)(1e9 + 7);
const int maxn = 500 * 1000 + 100;
clock_t cl;
double current_time() { return double(clock() - cl) / CLOCKS_PER_SEC; }
//------------------------------------------//

int mat[1000][1000];
bool check(int n, vector<int>& lst){
    int idx = 0;
    for (int i = 0; i < n; ++i){
        for (int j = 0; j < n; ++j){
            mat[i][j] = 0;
        }
    }
    for (int i = 1; i < n && idx < (int)lst.size(); i += 2){
        for (int j = 0; j < n && idx < (int)lst.size(); j += 2){
            mat[i][j] = lst[idx++];
        }
    }
    for (int i = 0; i < n && idx < (int)lst.size(); i += 2){
        for (int j = 0; j < n && idx < (int)lst.size(); j+=2){
            mat[i][j] = lst[idx++];
        }
    }
    for (int i = 0; i < n && idx < (int)lst.size(); i += 2){
        for (int j = 1; j < n && idx < (int)lst.size(); j+=2){
            mat[i][j] = lst[idx++];
        }
    }
    //cout << n << endl;
    //for (int i = 0; i <  n; ++i){
    //    for (int j = 0;  j< n; ++j){
    //        cout << mat[i][j] << " ";
    //    }
    //    cout << endl;
    //}
    if (idx < (int)lst.size()) return false;
    for (int i = 0; i < n-1; ++i){
        for (int j = 0; j < n-1; ++j){
            if (mat[i][j] == 0 && mat[i+1][j+1] == 0) continue;
            if (mat[i+1][j] == 0 && mat[i][j+1] == 0) continue;
            if (mat[i][j] == mat[i+1][j+1]) return false;
            if (mat[i+1][j] == mat[i][j+1]) return false;
        }
    }
    return true;

}

void solve(){
    int m, k;
    cin >> m >> k;
    map<int,int> ma;
    for (int i = 1; i <= k; ++i) 
        cin >> ma[i];
    vector<pair<int,int>> vp;
    for (auto& it : ma) vp.push_back(mp(it.S, it.F));
    sort(all(vp));
    reverse(all(vp));
    vector<int> lst;
    for (auto& it : vp) {
        for (int i =0 ; i < it.F; ++i)
            lst.push_back(it.S);
    }
    int l = 1;
    int r = sqrt(3.5*m)+10;
    while(l < r){
        int mid = (l+r)/2;
        if (check(mid, lst) == false) l = mid+1;
        else r = mid;
    }
    check(r,lst);
    cout << r << endl;
    for (int i = 0; i < r; ++i){
        for (int j = 0; j < r; ++j){
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }
}



int32_t main(){
    cl = clock();
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    cin.tie(nullptr);
#ifdef _MY
    freopen("VimProject/input.txt", "r", stdin);
    freopen("VimProject/output.txt", "w", stdout);
#endif

    int t;
    cin >> t;
    while(t--) solve();


    return 0;
}