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>
using namespace std;
#define int long long

const int LIM = 1e4, INF = 1e18;

signed main(){
	cin.tie(0)->sync_with_stdio(0);
	int T; cin >> T;
	while(T--){
		int n, m; cin >> n >> m;

		int res = 0;
		int y[m];

		int x[n]; for(int &i : x) cin >> i;

		bitset<LIM> s[n];
		for(auto &i : s) cin >> i;

		for(int i=0; i<(1<<n); ++i){
			int cost[m] = {}, cnt[m] = {};

			for(int j=0; j<n; ++j){
				int add = (i & (1<<j)) ? -1 : 1;
				for(int k=0; k<m; ++k)
					if(s[j][k]) cost[k] += add, cnt[k] += add * x[j];
			}

			int curr = 0;

			int b[m]; iota(b, b+m, 0LL);
			sort(b, b+m, [&](int i, int j){
				return make_pair(cost[i], cnt[i]) < make_pair(cost[j], cnt[j]);
			});

			for(int k=0; k<m; ++k)
				cost[b[k]] = k+1;

			for(int j=0; j<n; ++j){
				int sum = 0;
				for(int k=0; k<m; ++k)
					if(s[j][k]) sum += cost[k];
				curr += abs(x[j] - sum);
			}

			if(curr > res){
				res = curr;
				for(int k=0; k<m; ++k)
					y[k] = cost[k];
			}
		}

		reverse(y, y+m);
		for(int &i : y) cout << i << ' ';
		cout << '\n';
	}
}