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 math import ceil
import heapq

def solve(m,k,a):
    N = 1
    free = ceil(N/2)**2
    A = (N//2)*ceil(N/2)
    B = (N//2)*ceil(N/2)

    while free + A + B < m:
        N += 1
        free = ceil(N/2)**2
        A = (N//2)*ceil(N/2)
        B = (N//2)*ceil(N/2)

    mx_freq = max(a)
    while free + B < mx_freq:
        N += 1
        free = ceil(N/2)**2
        A = (N//2)*ceil(N/2)
        B = (N//2)*ceil(N/2)
    
    ans = [[0]*N for _ in range(N)]

    A = [(i,j) for i in range(1,N,2) for j in range(0,N,2)]
    B = [(i,j) for i in range(0,N,2) for j in range(1,N,2)]
    F = [(i,j) for i in range(0,N,2) for j in range(0,N,2)]

    pq = []
    for i in range(k):
        if a[i] > 0:
            heapq.heappush(pq,(-a[i],i+1))
    l = len(pq)
    cur_ele = None
    cur_freq = 0
    while B:
        if not pq and cur_freq == 0:
            break

        if cur_freq == 0:
            X = heapq.heappop(pq)
            cur_ele = X[1]
            cur_freq = -X[0]

        i,j = B.pop()
        ans[i][j] = cur_ele
        cur_freq -= 1
    
    later = []
    if len(pq) == l-1:
        later.append([cur_ele,cur_freq])
        cur_ele = None
        cur_freq = 0
    while A:
        if not pq and cur_freq == 0:
            break

        if cur_freq == 0:
            X = heapq.heappop(pq)
            cur_ele = X[1]
            cur_freq = -X[0]

        i,j = A.pop()
        ans[i][j] = cur_ele
        cur_freq -= 1
    
    if later:
        ele,freq = later.pop()
        if cur_freq:
            heapq.heappush(pq,(-cur_freq,cur_ele))
        cur_ele = ele
        cur_freq = freq
    while pq or cur_freq:
        if cur_freq == 0:
            X = heapq.heappop(pq)
            cur_ele = X[1]
            cur_freq = -X[0]

        while cur_freq:
            i,j = F.pop()
            ans[i][j] = cur_ele
            cur_freq -= 1
    
    print(N)
    for row in ans:
        print(*row)

    return -1
for nt in range(int(input())):
    m,k = map(int,input().split())
    a = list(map(int,input().split()))

    solve(m,k,a)