#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
pll operator+(pll p1, pll p2) {
	return pll(p1.first + p2.first, p1.second + p2.second);
}
pll dp[20000];
int mem[20000];
string ss = ".XO";
int chkv[100][3] = { {0, 1, 2}, {3, 4, 5}, {6, 7, 8}, {0, 3, 6}, {1, 4, 7}, {2, 5, 8}, {0, 4, 8}, {2, 4, 6} };
int pow3[100] = { 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683 };
int chk(string s) {
	int i;
	for (i = 0; i < 8; i++) {
		if (s[chkv[i][0]] == s[chkv[i][1]] && s[chkv[i][1]] == s[chkv[i][2]]) {
			if (s[chkv[i][0]] == '.') continue;
			if (s[chkv[i][0]] == 'X') return 1;
			if (s[chkv[i][0]] == 'O') return 2;
		}
	}
	return 0;
}
pll solve(int v, string s) {
	if (mem[v]) return dp[v];
	mem[v] = 1;
	int res = chk(s);
	if (res == 1) return dp[v] = pll(1, 0);
	if (res == 2) return dp[v] = pll(0, 1);
	int i;
	int x, o;
	x = o = 0;
	for (i = 0; i < 9; i++) {
		if (s[i] == 'O') o++;
		if (s[i] == 'X') x++;
	}
	int t = 0;
	if (x <= o) t = 1;
	else t = 2;

	for (i = 0; i < 9; i++) {
		if (s[i] != '.') continue;
		s[i] = ss[t];
		dp[v] = dp[v] + solve(v + pow3[i] * t, s);
		s[i] = '.';
	}
	return dp[v];
}
signed main() {
	ios::sync_with_stdio(false), cin.tie(0);
	solve(0, string(9, '.'));
	int T;
	cin >> T;
	while (T--) {
		string s;
		cin >> s;
		int i;
		int sum = 0;
		for (i = 0; i < 9; i++) {
			if (s[i] == 'X') sum += pow3[i];
			if (s[i] == 'O') sum += pow3[i] * 2;
		}
		if (!mem[sum]) cout << "-1 -1\n";
		else cout << dp[sum].first << ' ' << dp[sum].second << '\n';
	}
}