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 <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <cstring>
#include <string>
#include <cmath>
#include <cassert>
#include <ctime>
#include <algorithm>
#include <sstream>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <iterator>
#include <functional>
#include <bitset>
#include <iomanip>
#include <float.h>
#include <numeric>
#define _USE_MATH_DEFINES
#include <math.h>

#define ll long long

#define FOR(i,a,b)        for(ll i=(a);i<(b);i++)
#define REP(i,n)        FOR(i,0,n)
#define SORT(v)            sort((v).begin(),(v).end())
#define UN(v)            SORT(v),(v).erase(unique((v).begin(),(v).end()),(v).end())
#define CL(a,b)            memset(a,b,sizeof a)
#define pb                push_back


#ifndef M_PI
    #define M_PI 3.14159265358979323846
#endif

using namespace std;

ll denom = 1e9+7;//998244353;
ll powmod(ll a,ll b) {ll res=1;a%=denom; for(;b;b>>=1){if(b&1)res=res*a%denom;a=a*a%denom;}return res;}

bool fit(ll side, ll maxSingle, ll total) {
    if (side*((side+1)/2) < maxSingle) return false;
    return side*((side+1)/2) + (side/2)*((side+1)/2) >= total;
}

ll next(vector<pair<ll, ll>>& a) {
    ll last = a[a.size()-1].second+1;
    a[a.size()-1].first--;
    while (a.size() > 0 && a[a.size()-1].first == 0) a.pop_back();
    return last;
}

int main(int argc, const char * argv[])
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    ll t; cin >> t;
    while (t-- > 0) {
        ll m,k; cin >> m >> k;
        vector<pair<ll, ll>> a(k); REP(i, k) { cin >> a[i].first; a[i].second = i; }
        ll sum = 0; SORT(a); for (auto p : a) sum += p.first;
        ll l = 1; ll h = 1e6;
        while (l < h) {
            ll mid = (l+h)/2;
            fit(mid, a[k-1].first, sum) ? h = mid : l = mid+1;
        }
        cout << l << endl;
        vector<vector<ll>> result(l, vector<ll>(l, 0));
        
        for (ll i = 0; i < l && a.size() > 0; i+= 2) for (ll j = 1; j < l && a.size() > 0; j+= 2) result[i][j] = next(a);
        for (ll i = 0; i < l && a.size() > 0; i+= 2) for (ll j = 0; j < l && a.size() > 0; j+= 2) result[i][j] = next(a);
        for (ll i = 1; i < l && a.size() > 0; i+= 2) for (ll j = 0; j < l && a.size() > 0; j+= 2) result[i][j] = next(a);
        
        REP(i, l) { REP(j, l) cout << result[i][j] << " "; cout << endl; }
    }
    return 0;
}