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;

#define dbg(x...) \
	do { \
		cout << #x << " -> "; \
		err(x); \
	} while (0)

void err() {
    cout << endl;
}

template<class T, class... Ts>
void err(T arg, Ts &... args) {
    cout << arg << ' ';
    err(args...);
}

void read() {}

template <class T, class... Ts>
void read(T &arg, Ts &... args) {
    T f = 1;
    char ch;
    arg = 0;
    for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar()) {
        if (ch == '-') f = -1;
    }
    for (; ch >= '0' && ch <= '9'; ch = getchar()) arg = arg * 10 + ch - '0';
    arg *= f;
    read(args...);
}

template <class T>
void myHash(T a[], int n, T w[] = nullptr) {
    set<T> st;
    for (int i = 0; i < n; i++) {
        st.insert(a[i]);
    }
    int tot = 0;
    map<T, int> mp;
    for (T x : st) {
        mp[x] = ++tot;
    }
    for (int i = 0; i < n; i++) {
        a[i] = mp[a[i]];
    }
    if (w != nullptr) {
        for (pair<T, int> p : mp) {
            w[p.second] = p.first;
        }
    }
}

typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;

int a[maxn];

bool check(int n, int m, int mx) {
    int x1 = n / 2;
    int x2 = (n + 1) / 2;
    return m <= n * n - x1 * x1 && mx <= n * x2;
}

int f[1005][1005];

struct Cell {
    int x, y, type;
    bool operator < (const Cell &tmp) const {
        return type < tmp.type;
    }
};

void run() {
    int m, k;
    read(m, k);
    int mx = 0;
    for (int i = 1; i <= k; i++) {
        read(a[i]);
        mx = max(mx, a[i]);
    }
    int l = 1, r = 1000;
    int n = 0;
    while (l <= r) {
        int mid = l + r >> 1;
        if (check(mid, m, mx)) {
            n = mid;
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }
    printf("%d\n", n);

    vector<pii> vec;
    for (int i = 1; i <= k; i++) {
        vec.emplace_back(a[i], i);
    }
    sort(vec.rbegin(), vec.rend());

    vector<Cell> cell;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            f[i][j] = 0;
            if (i % 2 == 0 && j % 2 == 0) continue;
            if (j % 2 == 0) {
                cell.push_back({i, j, 1});
            } else if (i & 1) {
                cell.push_back({i, j, 2});
            } else {
                cell.push_back({i, j, 3});
            }
        }
    }
    sort(cell.begin(), cell.end());
    int cur = 0;
    for (pii p : vec) {
        int x = p.second;
        int cnt = p.first;
        while (cnt--) {
            Cell pos = cell[cur];
            cur++;
            f[pos.x][pos.y] = x;
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            printf("%d%c", f[i][j], j == n ? '\n' : ' ');
        }
    }
}

/*
5
1 2 3 4
*/

int main() {

    int T = 1;
    read(T);
    while (T--) {
        run();
    }

    return 0;
}