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<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<map>
#include<vector>
#include<set>

using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<ll> vl;
#define eb emplace_back
#define all(x) begin(x), end(x)
#define FOR(i, a, b) for(ll i=(a); i<=(b); i++)
#define ROF(i, a, b) for(ll i=(a); i>=(b); i--)
#define chkmax(x, a) x = max(x, a)
#define chkmin(x, a) x = min(x, a)

ll fillable, any, na, nb;
void insert(queue<ll>& a, queue<ll>& x, ll ts, ll ti, ll dstSz) {
    FOR(i, 1, ts) {
        if (a.size() != dstSz) a.emplace(ti);
        else x.emplace(ti);
    }
}
void solve(){
    ll n, nc; cin >> n >> nc; ll t;
    vector<pair<ll, ll>> c; FOR(i, 1, nc) cin >> t, c.eb(t, i);
    sort(all(c), greater<pair<ll, ll>>());
    ll lo = 1, hi = 1000;
    while (lo < hi) {
        ll mi = lo + hi >> 1;
        fillable = mi * mi - (mi / 2) * (mi / 2);
        any = ((mi + 1) / 2) * ((mi + 1) / 2);
        na = nb = ((mi + 1) / 2) * (mi / 2);
        if (fillable >= n && na + any >= c[0].first) hi = mi;
        else lo = mi + 1;
    }
    fillable = lo * lo - (lo / 2) * (lo / 2);
    any = ((lo + 1) / 2) * ((lo + 1) / 2);
    na = nb = ((lo + 1) / 2) * (lo / 2);
    vector<vl> m(lo + 1, vl(lo+1));
    queue<ll> a, b, x;
    for (auto [ts, ti]: c) {
        if (a.size() != na) insert(a, x, ts, ti, na);
        else insert(b, x, ts, ti, nb);
    }
    cout << lo << endl;
    FOR(i, 1, lo) {
        FOR(j, 1, lo) {
            if (i % 2 && j % 2 && x.size()) m[i][j] = x.front(), x.pop();
            else if (i % 2 && j % 2 == 0 && a.size()) m[i][j] = a.front(), a.pop();
            else if (i % 2 == 0 && j % 2 && b.size()) m[i][j] = b.front(), b.pop();

            cout << m[i][j] << " ";
        }
        cout << "\n";
    }
    cout.flush();
}

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

    ll T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
}