Petya is a math teacher. n of his students has written a test consisting of m questions. For each student, it is known which questions he has answered correctly and which he has not.

If the student answers the j-th question correctly, he gets p_j points (otherwise, he gets 0 points). Moreover, the points for the questions are distributed in such a way that the array p is a permutation of numbers from 1 to m.

For the i-th student, Petya knows that he expects to get x_i points for the test. Petya wonders how unexpected the results could be. Petya believes that the surprise value of the results for students is equal to ∑_{i=1}^{n} |x_i - r_i|, where r_i is the number of points that the i-th student has got for the test.

Your task is to help Petya find such a permutation p for which the surprise value of the results is maximum possible. If there are multiple answers, print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first line of each test case contains two integers n and m (1 ≤ n ≤ 10; 1 ≤ m ≤ 10^4) — the number of students and the number of questions, respectively.

The second line contains n integers x_1, x_2, ..., x_n (0 ≤ x_i ≤ (m(m+1))/(2)), where x_i is the number of points that the i-th student expects to get.

This is followed by n lines, the i-th line contains the string s_i (|s_i| = m; s_{i, j} ∈ \{0, 1\}), where s_{i, j} is 1 if the i-th student has answered the j-th question correctly, and 0 otherwise.

The sum of m for all test cases does not exceed 10^4.

Output

For each test case, print m integers — a permutation p for which the surprise value of the results is maximum possible. If there are multiple answers, print any of them.

Example

Input


3
4 3
5 1 2 2
110
100
101
100
4 4
6 2 0 10
1001
0010
0110
0101
3 6
20 3 15
010110
000101
111111


Output


3 1 2 
2 3 4 1 
3 1 4 5 2 6 
#include <bits/stdc++.h>
#include <ext/rope>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#define int long long
#define pb push_back
#define x first
#define y second
#define mk(a,b) make_pair(a,b)
#define rr return 0
#define sqr(a) ((a)*(a))
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#undef M_PI
#define M_PI 3.14159265358979323846264338327950288419

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using big = __int128;

template<class value, class cmp = less<value> >
using ordered_set = tree<value, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;
template<class value, class cmp = less_equal<value> >
using ordered_multiset = tree<value, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;
template<class key, class value, class cmp = less<key> >
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;

/// find_by_order()
/// order_of_key()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<typename T = int>
inline T randll(T l = INT_MIN, T r = INT_MAX) {
    return uniform_int_distribution<T>(l, r)(rng);
}

inline ld randld(ld l = INT_MIN, ld r = INT_MAX) {
    return uniform_real_distribution<ld>(l, r)(rng);
}

const int INF = 2e9 + 1;
const int MOD = 998244353; /// think
const ll LINF = 2e18 + 11;

const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
inline bool inside (int x, int y, int n, int m) {
    return 0 <= x && 0 <= y && x < n && y < m;
}

template<class T> bool umin (T &a, T b) {return a > b ? (a = b, true) : false; }
template<class T> bool umax (T &a, T b) {return a < b ? (a = b, true) : false; }


inline void test_case() {
    int n, m;
    cin >> n >> m;
    vector <int> x(n);
    for (auto &i : x) cin >> i;
    vector <string> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    ll ans = -1;
    vector <int> res(m);
    int cnt[21] = {};
    vector <vector <int> > pos(21);
    vector <int> p(m);
    for (int mask = 0; mask < (1 << n); mask++) {
//        memset(cnt, 0, sizeof(cnt));
        for (int j = 0; j < m; j++) {
            int r = 0;
            for (int i = 0; i < n; i++) {
                if (a[i][j] == '1') {
                    if (mask & (1 << i)) {
                        ++r;
                    }
                    else --r;
                }
            }
            ++cnt[r + 10];
            pos[r + 10].pb(j);
        }
        ll cur = 0;
        int l = m;
        for (int i = 20; i >= 0; i--) {
            while (cnt[i] > 0) {
                --cnt[i];
                cur += l * (i - 10);
                --l;
                p[pos[i].back()] = l;
                pos[i].pop_back();
            }
        }
        if (umax(ans, cur)) {
            res.swap(p);
        }
    }
    for (auto &i : res) {
        cout << i + 1 << ' ';
    }
    cout << '\n';
}

main()
{
    ios::sync_with_stdio(0);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    for (int test = 1; test <= t; test++) {
        test_case();
    }
}
