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;

using ll = long long;
using ld = long double;

#define vec vector
#define ar array

#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define len(a) ((int)((a).size()))

template<typename A, typename B>
ostream& operator<<(ostream &out, pair<A, B> p) { return out << '(' << p.first << ", " << p.second << ')'; }

template<typename A, typename B = typename enable_if<!is_same<A, string>::value, typename A::value_type>::type>
ostream& operator<<(ostream &out, A vec) { out << '{'; string sep; for (B x : vec) out << sep << x, sep = ", "; return out << '}'; }

void dbg_print() { cerr << endl; }
template<typename A, typename... B> void dbg_print(A a, B... b) { cerr << ' ' << a; dbg_print(b...); }

#ifdef LOCAL
    #define dbg(...) cerr << '#' << __LINE__ << " [" << #__VA_ARGS__ << "]:", dbg_print(__VA_ARGS__)
#else
    #define dbg(...)
#endif

void solve() {
    int m, k;
    cin >> m >> k;
    vec<int> cnt(k);
    for (auto &x : cnt) {
        cin >> x;
    }
    vec<int> sorted(k);
    iota(all(sorted), 0);
    sort(all(sorted), [&](int a, int b) { return cnt[a] > cnt[b]; });
    auto solve = [&](int n) {
        vec<pair<int, int>> ord;
        for (int i = 0; i < n; i += 2) {
            for (int j = 1; j < n; j += 2) {
                ord.emplace_back(i, j);
            }
        }
        for (int i = 0; i < n; i += 2) {
            for (int j = 0; j < n; j += 2) {
                ord.emplace_back(i, j);
            }
        }
        for (int i = 1; i < n; i += 2) {
            for (int j = 0; j < n; j += 2) {
                ord.emplace_back(i, j);
            }
        }
        if (len(ord) < m) return vec<vec<int>>();
        vec<vec<int>> mat(n, vec<int>(n));
        int pos = 0;
        for (auto i : sorted) {
            for (int j = 0; j < cnt[i]; j++) {
                mat[ord[pos].first][ord[pos].second] = i + 1;
                pos++;
            }
        }
        return mat;
    };
    int l = 0, r = min(1000, m);
    while (r - l > 1) {
        int mid = (l + r) / 2;
        auto mat = solve(mid);
        bool ok = (len(mat) >= 1);
        for (int i = 0; i < len(mat) && ok; i++) {
            for (int j = 0; j < len(mat) && ok; j++) {
                if (i >= 1 && j >= 1 && mat[i][j] == mat[i - 1][j - 1] && mat[i][j] != 0) ok = false;
                if (i < len(mat) - 1 && j > 0 && mat[i][j] == mat[i + 1][j - 1] && mat[i][j] != 0) ok = false;
            }
        }
        if (ok) {
            r = mid;
        } else {
            l = mid;
        }
    }
    auto mat = solve(r);
    cout << r << '\n';
    for (auto xs : mat) {
        for (auto x : xs) {
            cout << x << ' ';
        }
        cout << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

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