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;
typedef pair<int, int> pii;
int n;
int a[100005];
int b[2005][2005];
bool vis[2005][2005];

int cnt[2][2];
void dfs(int i, int j, int k, int l){
    cnt[k][l]++;
    vis[i][j] = 1;
    for(int x=-1;x<=1;x+=2){
        for(int y=-1;y<=1;y+=2){
            int x2 = i+x, y2 = j+y;
            if(1 <= x2 && x2 <= n && 1 <= y2 && y2 <= n){
                if(vis[x2][y2]) continue;
                dfs(x2, y2, k, (l+1)%2);
            }
        }
    }
}

vector<pii> v[2][2];
void dfs2(int i, int j, int k, int l){
    cnt[k][l]++;
    v[k][l].emplace_back(make_pair(i, j));
    vis[i][j] = 1;
    for(int x=-1;x<=1;x+=2){
        for(int y=-1;y<=1;y+=2){
            int x2 = i+x, y2 = j+y;
            if(1 <= x2 && x2 <= n && 1 <= y2 && y2 <= n){
                if(vis[x2][y2]) continue;
                dfs2(x2, y2, k, (l+1)%2);
            }
        }
    }
}

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        int m, k;
        scanf("%d%d", &m, &k);
        for(int i=1;i<=k;i++) scanf("%d", &a[i]);

        if(m == 1){
            printf("1\n");
            for(int i=1;i<=k;i++){
                if(a[i] == 1){
                    printf("%d\n", i);
                }
            }
            continue;
        }

        int sq = (sqrt(m)+5)*2;
        int lo = 2, hi = sq, ans;
        while(lo <= hi){
            int mid = (lo+hi)/2;
            n = mid;
            for(int i=1;i<=mid;i++){
                for(int j=1;j<=mid;j++){
                    vis[i][j] = 0;
                }
            }

            priority_queue<int> pq;
            for(int i=1;i<=k;i++) pq.push(a[i]);

            cnt[0][0] = 0;
            cnt[0][1] = 0;

            cnt[1][0] = 0;
            cnt[1][1] = 0;

            dfs(1, 2, 0, 0);
            dfs(1, 1, 1, 0);

            int x = 0, y = 0;
            if(cnt[0][0] < cnt[0][1]) x = 1;
            if(cnt[1][0] < cnt[1][1]) y = 1;

            bool ok = 1;
            while(!pq.empty()){
                int val = pq.top();
                pq.pop();
                if(cnt[0][x^1] >= cnt[0][x]) x ^= 1;
                if(cnt[0][x] >= val){
                    cnt[0][x] -= val;
                }
                else{
                    val -= cnt[0][x];
                    cnt[0][x] = 0;
                    if(cnt[1][y] < val){
                        ok = 0;
                        break;
                    }
                    cnt[1][y] -= val;
                }
            }
            if(ok) ans = mid, hi = mid-1;
            else lo = mid+1;
        }

        printf("%d\n", ans);
        for(int i=0;i<2;i++){
            for(int j=0;j<2;j++) v[i][j].clear();
        }
        for(int i=1;i<=ans;i++){
            for(int j=1;j<=ans;j++){
                vis[i][j] = 0;
            }
        }

        n = ans;
        cnt[0][0] = 0;
        cnt[0][1] = 0;

        cnt[1][0] = 0;
        cnt[1][1] = 0;

        dfs2(1, 2, 0, 0);
        dfs2(1, 1, 1, 0);

        priority_queue<pii> pq;
        for(int i=1;i<=k;i++){
            pq.push(make_pair(a[i], i));
        }

        int x = 0, y = 0;
        if(cnt[0][0] < cnt[0][1]) x = 1;
        if(cnt[1][0] < cnt[1][1]) y = 1;

        for(int i=1;i<=ans;i++){
            for(int j=1;j<=ans;j++) b[i][j] = 0;
        }

        bool ok = 1;
        while(!pq.empty()){
            int val = pq.top().first;
            int idx = pq.top().second;
            pq.pop();
            if(cnt[0][x^1] >= cnt[0][x]) x ^= 1;
            if(cnt[0][x] >= val){
                cnt[0][x] -= val;
                while(!v[0][x].empty() && val > 0){
                    int l = v[0][x].back().first, r = v[0][x].back().second;
                    v[0][x].pop_back();
                    b[l][r] = idx;
                    val--;
                }
            }
            else{
                cnt[0][x] = 0;
                while(!v[0][x].empty() && val > 0){
                    int l = v[0][x].back().first, r = v[0][x].back().second;
                    v[0][x].pop_back();
                    b[l][r] = idx;
                    val--;
                }
                if(cnt[1][y] < val){
                    ok = 0;
                    break;
                }
                cnt[1][y] -= val;
                while(!v[1][y].empty() && val > 0){
                    int l = v[1][y].back().first, r = v[1][y].back().second;
                    v[1][y].pop_back();
                    b[l][r] = idx;
                    val--;
                }
            }
        }
        for(int i=1;i<=ans;i++){
            for(int j=1;j<=ans;j++) printf("%d ", b[i][j]);
            printf("\n");
        }
    }
}