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>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
#define DBG(a) cerr<< "line "<<__LINE__ <<" : "<< #a <<" --> "<<(a)<<endl
#define eps 1e-8
#define eq(x,y) (fabs((x)-(y)) < eps)
using namespace std;
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int>pii;
int mod = 1e9+7;
long double pi = acosl(-1);
const ll infl = 1e17+100;
const int inf = 1e9+100;
const int nmax = 1e5+5;
const int MAXLG = log2(nmax)+1;
//mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
//typedef tree< ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> ost;


int m,k;
vector< vector<int> >arr;
vector<pii>v;
bool cmp(pii a, pii b){
    if(a.first%2 == b.first%2) return a < b;
    return a.first%2 > b.first%2;
}

bool chek(int n){

    int bhalo =  ((n+1)/2) * ((n+1)/2);
    int kharap = n*(n+1)/2 - (n+1)/2;

    if(m > bhalo + kharap) return false;
    int beshi = v[0].first;
    if( beshi-bhalo-kharap/2 > 0 ) return false;

    arr.resize(n);
    for(int i=0; i<n; i++) arr[i] = vector<int>(n, 0);
    vector<pii>gudpos;
    vector<pii>bad1;
    vector<pii>bad2;


    for(int i=0; i<n; i++) for(int j=0; j<n; j++){
        if(i%2 and j%2) continue;
        if(i%2==0 and j%2==0) gudpos.push_back({i,j});
        else {
            if(i%2)
                bad1.push_back({i,j});
            else
                bad2.push_back({i,j});
        }
    }
    vector<pii>serial;
    for(pii z : bad1) serial.push_back(z);
    for(pii z : gudpos) serial.push_back(z);
    for(pii z : bad2) serial.push_back(z);

    int cur = 0;
    for(pii z : v) if(z.first) {
        int koyta = z.first;
        int num = z.second;
        for(int j=0; j<koyta; j++){
            arr[ serial[cur].first ][ serial[cur].second ] = num;
            cur++;
        }
    }
    return true;
}


void solv(){
    cin>>m>>k;
    v.resize(k);
    for(int i=0; i<k; i++) v[i].second = i+1;
    for(pii &x : v) cin>>x.first;
    sort(v.rbegin(), v.rend());

    int lo = 1;
    while(!chek(lo)) lo++;
    cout<<lo<<"\n";
    for(int i=0; i<lo; i++){
        for(int j=0; j<lo; j++){
            cout<<arr[i][j]<<" ";
        }
        cout<<"\n";
    }
}

int main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);


    int tc;
    cin>>tc;
    while(tc--){
        solv();
    }
}
/*

*/