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>
#define ll long long
#define f first
#define s second
#define pb push_back
using namespace std;
ll m,k,c[2005][2005];
pair<ll,ll>a[500005],b[500005];
bool check(ll x,ll n){
    if((x + 1) / 2 * x < a[1].f)return 0;
    ll sr = (x + 1) / 2;
    ll nr = x - sr;
    ll num = sr * x + nr * ((x + 1) / 2);
    return (num >= n);
}
void solve(){
    cin >> m >> k;
    for(int i=1; i<=k; i++){
        cin >> a[i].f;
        a[i].s = i;
        b[i] = a[i];
    }
    sort(b + 1 , b + k + 1);
    reverse(b + 1 , b + k + 1);
    sort(a + 1 , a + k + 1);
    reverse(a + 1 , a + k + 1);
    ll l = 0 , r = 1e5,mid,ans = 0;
    while(r >= l){
        mid = (l + r) / 2;
        if(check(mid , m)){
            r = mid - 1;
            ans = mid;
        }
        else {
            l = mid + 1;
        }
    }
    while(1){
        for(int i=1; i<=k; i++)
            a[i] = b[i];
        ll ind = 1;
        for(int i=1; i<=ans; i+=2){
            for(int j=1; j<=ans; j++){
                while(!a[ind].f && ind <= k)ind++;
                if(ind > k)break;
                if(a[ind].f){
                    a[ind].f--;
                    c[i][j] = a[ind].s;
                }
            }
            if(ind > k)break;
        }
        for(int i=2; i<=ans; i+=2){
            for(int j=1; j<=ans; j+=2){
                while(!a[ind].f && ind <= k)ind++;
                if(ind > k)break;
                if(a[ind].f){
                    a[ind].f--;
                    c[i][j] = a[ind].s;
                }
            }
            if(ind > k)break;
        }
        for(int i=1; i<=k; i++)
            a[i] = {0LL , 0LL};
        bool ye = 0;
        for(int i=1; i<ans; i++){
            for(int j=1; j<ans; j++){
                if(c[i][j] && c[i][j] == c[i + 1][j + 1]){
                    ye = 1;
                }
            }
        }
        for(int i=1; i<=ans; i++){
            for(int j=2; j<=ans; j++){
                if(c[i][j] && c[i][j] == c[i + 1][j - 1]){
                    ye = 1;
                }
            }
        }
        if(!ye){
            break;
        }
        else{
            l = ans;
            ll raod = 0;
            if(ans % 2 == 0)l--;
            ll las = c[l][ans];
            ll indx = 1,indy = 1;
            for(int l=1; l<=ans; l+=2)
            for(int i=2; i<=ans; i+=2){
                if(c[l][i] == las){
                    if(c[l][i] != c[indx][indy]){
                        swap(c[indx][indy] , c[l][i]);
                        indy += 2;
                        if(indy > ans){
                            indy = 1;
                            indx += 2;
                        }
                        if(indx > ans)break;
                    }
                    else{
                        indy += 2;
                        if(indy > ans){
                            indy = 1;
                            indx += 2;
                        }
                        if(indx > ans)break;
                    }
                }
            }
            break;
        }
    }
    bool ye = 0;
    for(int i=1; i<ans; i++){
        for(int j=1; j<ans; j++){
            if(c[i][j] && c[i][j] == c[i + 1][j + 1]){
                ye = 1;
            }
        }
    }
    for(int i=1; i<=ans; i++){
        for(int j=2; j<=ans; j++){
            if(c[i][j] && c[i][j] == c[i + 1][j - 1]){
                ye = 1;
            }
        }
    }
    if(ye){
        while(true);
    }
    cout << ans << '\n';
    for(int i=1; i<=ans; i++){
        for(int j=1; j<=ans; j++){
            cout << c[i][j] << " ";
        }
        cout << '\n';
    }
    for(int i=1; i<=ans; i++)
        for(int j=1; j<=ans; j++)
            c[i][j] = 0;
}
int main(){
    ios::sync_with_stdio(false);
    ll q;
    cin >> q;
    
    while(q--){
        solve();
    }
    
    
    
    return 0;
}