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 long long ll;
typedef long double ld;

typedef vector<int> vi;
typedef vector<ll> vll;

typedef vector<vi> vvi;
typedef vector<vll> vvll;

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

typedef vector<pii> vpii;
typedef vector<pll> vpll;

typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()
#define fi first
#define se second

template<class T> bool ckmin(T &a, const T &b) {return a > b ? a = b, 1 : 0;}
template<class T> bool ckmax(T &a, const T &b) {return a < b ? a = b, 1 : 0;}

namespace debug {
    void __print(int x) {cerr << x;}
    void __print(long long x) {cerr << x;}
    void __print(double x) {cerr << x;}
    void __print(long double x) {cerr << x;}
    void __print(char x) {cerr << '\'' << x << '\'';}
    void __print(const string &x) {cerr << '\"' << x << '\"';}
    void __print(bool x) {cerr << (x ? "true" : "false");}

    template<typename T, typename V>
    void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
    template<typename T>
    void __print(const T &x) {int f = 0; cerr << '{'; for(auto z : x) cerr << (f++ ? "," : ""), __print(z); cerr << "}";}
    void _print() {cerr << "]\n";}
    template <typename T, typename... V>
    void _print(T t, V... v) {__print(t); if(sizeof...(v)) cerr << ", "; _print(v...);}

#ifdef ljuba
#define dbg(x...) cerr << "LINE(" << __LINE__ << ") -> " << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif
}

using namespace debug;

const char nl = '\n';

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());

int mat[1000][1000];
int m, k;

bool moze(int n, const vi &v) {
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            mat[i][j] = -1;
        }
    }

    vpii nz;
    for(int i = 0; i < k; ++i) {
        if(v[i]) {
            nz.pb({v[i], i+1});
        }
    }
    sort(all(nz));

    for(int i = 1; i < n; i += 2) {
        for(int j = 0; j < n; j += 2) {
            if(mat[i][j] != -1) continue;
            if(sz(nz) && nz.back().fi > 0) {
                nz.back().fi--;
                mat[i][j] = nz.back().se;
                if(nz.back().fi == 0)
                    nz.pop_back();
            }
        }
    }

    pii kasnije{-1, -1};

    if(sz(nz) && nz.back().fi > 0) {
        kasnije = nz.back();
        nz.pop_back();
    }

    dbg(kasnije);

    for(int i = 0; i < n; i += 2) {
        for(int j = 1; j < n; j += 2) {
            if(mat[i][j] != -1) continue;
            if(sz(nz) && nz.back().fi > 0) {
                nz.back().fi--;
                mat[i][j] = nz.back().se;
                if(nz.back().fi == 0)
                    nz.pop_back();
            }
        }
    }

    if(kasnije.fi != -1)
        nz.pb(kasnije);

    for(int i = 0; i < n; i += 2) {
        for(int j = 0; j < n; j += 2) {
            if(mat[i][j] != -1) continue;
            if(sz(nz) && nz.back().fi > 0) {
                nz.back().fi--;
                mat[i][j] = nz.back().se;
                if(nz.back().fi == 0)
                    nz.pop_back();
            }
        }
    }

    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            if(i % 2 == 1 && j % 2 == 1) continue;
            if(sz(nz)) {
                nz.back().fi--;
                mat[i][j] = nz.back().se;
                if(nz.back().fi == 0)
                    nz.pop_back();
            }
        }
    }

    return nz.empty();
}

void solve() {
    cin >> m >> k;
    vi v(k);
    for(auto &z : v) cin >> z;

    int mx = *max_element(all(v));
    int n = -1;
    for(int i = 1; ; ++i) {
        ll izraz = i * 1LL * ((i + 1) / 2);
        if(mx <= izraz) {
            ckmax(n, i);
            break;
        }
    }
    for(int i = 1; ; ++i) {
        ll izraz = i * 1LL * i;
        ll temp = i / 2;
        izraz -= temp * 1LL * temp;
        if(m <= izraz) {
            ckmax(n, i);
            break;
        }
    }

    while(!moze(n, v))
        ++n;

    cout << n << nl;
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            if(mat[i][j] == -1) mat[i][j] = 0;
            cout << mat[i][j] << " ";
        }
        cout << nl;
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

#ifdef ljuba
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
#endif

    int testCases = 1;
    cin >> testCases;
    while(testCases--)
        solve();

}