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 PYTHON3):

Solution:
from random import randint

DEBUG = 0

for _ in range(int(input())):
    if DEBUG:
        k = randint(2, 4)
        maxcnt = randint(1, 100)
        cnt = [randint(1, maxcnt) for i in range(k)]
        m = sum(cnt)
        #print(m, k)
        #print(*cnt)
        cnt = sorted(enumerate(cnt), key=lambda x: -x[1])
    else:
        m, k = list(map(int, input().split()))
        cnt = sorted(enumerate(map(int, input().split())), key=lambda x: -x[1])
    l = 0
    r = m
    while r - l > 1:
        mid = (r + l) // 2
        if mid * ((mid + 1) // 2) >= cnt[0][1] and (mid // 2) ** 2 + m <= mid ** 2:
            r = mid
        else:
            l = mid
    size = r
    queue = []
    #for it in range((size - 1) // 4 + 1):
        #queue.extend([(i, it * 4) for i in range(size)])
        #if it * 4 + 2 < size:
            #queue.extend([(i, it * 4 + 2) for i in range(1, size, 2)])
    #for it in range((size - 1) // 4 + 1):
        #if it * 4 + 2 < size:
            #queue.extend([(i, it * 4 + 2) for i in range(0, size, 2)])
    #queue.extend([(i, j) for i in range(0, size, 2) for j in range(1, size, 2)])
    #for it in range((size - 1) // 2 + 1):
        #queue.extend([(it * 2, j) for j in range(0, size, 4)])
        #if it * 2 + 1 < size:
            #queue.extend([(it * 2 + 1, j) for j in range(0, size, 2)])
    #queue.extend([(i, j) for i in range(0, size, 2) for j in range(2, size, 4)])
    #queue.extend([(i, j) for i in range(0, size, 2) for j in range(1, size, 2)])
    queue.extend([(i, j) for i in range(1, size, 2) for j in range(0, size, 2)])
    queue.extend([(i, j) for i in range(0, size, 2) for j in range(0, size, 4)])
    queue.extend([(i, j) for i in range(0, size, 2) for j in range(2, size, 4)])
    queue.extend([(i, j) for i in range(0, size, 2) for j in range(1, size, 2)])
    a = [[0] * size for i in range(size)]
    ind = 0
    rest = cnt[ind][1]
    for i, j in queue:
        while not rest:
            ind += 1
            if ind < k:
                rest = cnt[ind][1]
            else:
                break
        else:
            a[i][j] = cnt[ind][0] + 1
            rest -= 1
            continue
        break
    if not DEBUG:
        print(size)
        for row in a:
            print(*row)
    if DEBUG:
        for i in range(size - 1):
            for j in range(size - 1):
                if a[i][j] == a[i + 1][j + 1] != 0 or a[i + 1][j] == a[i][j + 1] != 0:
                    print("Oops...")
                    print(m, k)
                    print(*cnt)                    
                    break
            else:
                continue
            break
        else:
            continue
        break