#include <iostream>
#include <string>
#include <algorithm>
#define int long long
using namespace std;

int n;
string s[20];
int pre[20][20];	//pre[i][j] = 文字列s[i], s[j]の共通prefixの長さ
int dp[1 << 20];	//dp[i] = i…既に読んだ読み札の集合, dp[i]…そこまでの全読み方における疲労度の総和
int fact[20];

int getPreNum(string &s, string &t) {
	int n = min(s.length(), t.length());
	for (int i = 0; i < n; i++) {
		if (s[i] != t[i]) {
			return i;
		}
	}
	return n;
}

int bitCount(int x) {
	int cnt = 0;
	while (x > 0) {
		cnt += (x & 1);
		x >>= 1;
	}
	return cnt;
}

signed main() {
	int i, j, k;
	
	cin >> n;
	for (i = 0; i < n; i++) cin >> s[i];
	
	for (i = 0; i < n; i++) {
		for (j = 0; j < n; j++) {
			pre[i][j] = getPreNum(s[i], s[j]);
		}
	}
	
	fact[0] = 1;
	for (i = 1; i < n; i++) {
		fact[i] = fact[i - 1] * i;
		fact[i] %= 1000000007;
	}
	
	//遷移前 < 遷移後なので、値が小さい状態から調べればよい。
	for (i = 0; i < (1 << n) - 1; i++) {
		for (j = 0; j < n; j++) {
			if ((i >> j) & 1) continue;
			//読み札jを読む
			int tired = 0;
			for (k = 0; k < n; k++) {
				if (k != j && ((i >> k) & 1) == 0) {
					tired = max(tired, pre[j][k]);
				}
			}
			tired++;
			//遷移コスト = tired * その状態に至る経路の個数(=bitCount(i)!)
			dp[i + (1 << j)] += dp[i] + tired * fact[bitCount(i)];
			dp[i + (1 << j)] %= 1000000007;
		}
	}

	int ans[21] = {0};
	for (i = 0; i < (1 << n); i++) {
		ans[bitCount(i)] += dp[i];
		ans[bitCount(i)] %= 1000000007;
	}
	
	for (i = 1; i <= n; i++) {
		cout << ans[i] << endl;
	}
	return 0;
}