#include <bits/stdc++.h>
using namespace std;


class Solution {
public:
    inline bool delim(char c) { return c == ' '; }
    inline bool is_op(char c) { return c == '&' || c == '|' || c == '^' || c == '=' || c == '!'; }
    inline bool is_unary(char c) { return c == '!'; }
    inline int priority(char op) {
        if (op < 0) return 5; 
        if (op == '=') return 4; 
        if (op == '&') return 3; 
        if (op == '|') return 2; 
        if (op == '^') return 1; 
        return -1; 
    }

    void process_op(stack<int> &st, char op) {
        if (op < 0) {
            int l = st.top();
            st.pop();
            st.push(l ^ 1); 
        } else { 
            int r = st.top(); st.pop();
            int l = st.top(); st.pop(); 
            switch (op) {
            case '&':
                st.push(l & r);
                break;
            case '|':
                st.push(l | r);
                break;
            case '^':
                st.push(l ^ r);
                break;
            case '=':
                st.push(l == r);
                break;
            }
        }
    }

    int evaluate(string &s, int xval, int yval) {
        stack<int> st;
        stack<char> op;
        bool may_be_unary = true;
        for (int i = 0; i < int(s.size()); ++i) {
            if (delim(s[i])) continue;

            if (s[i] == '(') {
                op.push('('); // 2. 如果遇到左括号，那么将其放在运算符栈上
                may_be_unary = true;
            } else if (s[i] == ')') { // 3. 如果遇到右括号，执行一对括号内的所有运算符
                while (op.top() != '(') {
                    process_op(st, op.top());
                    op.pop(); // 不断输出栈顶元素，直至遇到左括号
                }
                op.pop(); // 左括号出栈
                may_be_unary = false;
            } else if (is_op(s[i])) { // 4. 如果遇到其他运算符
                char cur_op = s[i];
                if (may_be_unary && is_unary(cur_op))
                    cur_op = -cur_op;
                while (!op.empty() &&
                       ((cur_op >= 0 && priority(op.top()) >= priority(cur_op)) ||
                        (cur_op < 0 && priority(op.top()) > priority(cur_op)))) {
                    process_op(st, op.top());
                    op.pop(); // 不断输出所有运算优先级大于等于当前运算符的运算符
                }
                op.push(cur_op); // 新的运算符入运算符栈
                may_be_unary = true;
            } else { // 1. 如果遇到数字，直接输出该数字
                int number = 0;
                if (s[i] == 'x') number = xval; 
                else if (s[i] == 'y') number = yval; 
                else assert(isdigit(s[i])), number = s[i] - '0'; 
                st.push(number);
                may_be_unary = false;
            }
        }

        while (!op.empty()) {
            process_op(st, op.top());
            op.pop();
        }
        return st.top();
    }
} whiteqwq; 

int n, m, k; 
int ans[1024][1024]; 

struct Node {
    int x, y, val; 
    set<pair<int, int>> st; 
} a[200005]; 
int atval; 
vector<Node> t; 

inline int calc(int x, int y) {
    for (int i = 1; i <= m; ++i)
        if (a[i].st.find({x >> a[i].x & 1, y >> a[i].y & 1}) != a[i].st.end())
            return a[i].val; 
    return atval; 
}

set<tuple<int, int, int, int>> clc; 

int main(void) {
    ios::sync_with_stdio(0); 
    cin >> n >> k; 
    for (int i = 1; i <= n; ++i) {
        Node t; cin >> t.x >> t.y; --t.x; --t.y; 
        string s; cin >> s; 
        cin >> t.val; 
        if (whiteqwq.evaluate(s, 0, 0) && clc.find({t.x, t.y, 0, 0}) == clc.end()) t.st.emplace(0, 0), clc.emplace(t.x, t.y, 0, 0); 
        if (whiteqwq.evaluate(s, 0, 1) && clc.find({t.x, t.y, 0, 1}) == clc.end()) t.st.emplace(0, 1), clc.emplace(t.x, t.y, 0, 1); 
        if (whiteqwq.evaluate(s, 1, 0) && clc.find({t.x, t.y, 1, 0}) == clc.end()) t.st.emplace(1, 0), clc.emplace(t.x, t.y, 1, 0); 
        if (whiteqwq.evaluate(s, 1, 1) && clc.find({t.x, t.y, 1, 1}) == clc.end()) t.st.emplace(1, 1), clc.emplace(t.x, t.y, 1, 1); 
        if (t.st.size()) a[++m] = t; 
        cerr << i << "\n"; 
        for (auto [x, y] : t.st) cerr << x << " " << y << "\n"; 
        cerr << "----------\n"; 
    }
    cin >> atval; 

    for (int i = 0; i < 1 << k; ++i)
        for (int j = 0; j < 1 << k; ++j)
            ans[i][j] = calc(i, j); 
    cerr << calc(2, 2) << "\n"; 
    
    int ans1 = 0, ans2 = 0, ans3 = 0; 
    for (int i = 0; i < 1 << k; ++i) 
        if (ans[i][i] != 0) ++ans1; 
    for (int x = 0; x < 1 << k; ++x)
        for (int y = 0; y < 1 << k; ++y) {
            if (ans[x][y] == 1 && ans[y][x] != 0) ++ans2; 
            if (ans[x][y])
                for (int z = 0; z < 1 << k; ++z)
                    if (ans[y][z] && ans[x][z] == 0) ++ans3; 
        }
    cout << ans1 << " " << ans2 << " " << ans3 << "\n"; 
    return 0;
}