#include <ios>
#include <iostream>
#include <vector>

using namespace std;

struct ConvertILPToSat {
    vector< vector<int> > A;
    vector<int> b;

    ConvertILPToSat(int n, int m) : A(n, vector<int>(m)), b(n)
    {}

    void printEquisatisfiableSatFormula() {
        int vars = A[0].size();
        int eqs = A.size();
        vector<vector<int>> clauses;

        // 5x1 + 2x2 + 3x3 ≤ 6
        // vector that are no solutions :  (1,1,1)=10, (1,1,0)=7, (1,0,1)=8
        // corresponding clauses (!x || !y || !z), (!x || !y || z), (!x || y || !z)
        vector<int> coefficients(vars, -1);
        for (int i = 0; i < eqs; i++) {

            vector<int> equ = A[i];

            // collect non 0 coefficients
            coefficients.clear();
            for (int j = 0; j < vars; j++) {
                if (equ[j] != 0)  coefficients.push_back(j);
            }

            vector<int> clause;
            // for each bit combinaison
            for (int j = 0; j < (1<<coefficients.size()); j++) {

                // sum the coefficients using the bit map
                int v = 0;
                clause.clear();
                for (int k = 0; k < coefficients.size(); k++) {
                    bool b = ((j & (1<<k)) > 0);
                    int w = coefficients[k];
                    if (b) v += equ[w];
                    w += 1;
                    clause.push_back(b ? -w: w);
                }

                if (v > b[i]) {
                  clauses.push_back(clause);
                }
            }
        }

        int n = clauses.size();
        if (n == 0) {
            cout << "1 1\n1 -1 0"<< endl;
        } else {
            cout << clauses.size() << " " << vars << endl;
            for (vector<int>& c : clauses) {
                for (int v : c) {
                    cout << v << " ";
                }
                cout << "0" << endl;
            }
            cout << endl;
        }
    }
};

int main() {
    ios::sync_with_stdio(false);

    int n, m;
    cin >> n >> m;
    ConvertILPToSat converter(n, m);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        cin >> converter.A[i][j];
      }
    }
    for (int i = 0; i < n; i++) {
      cin >> converter.b[i];
    }

    converter.printEquisatisfiableSatFormula();

    return 0;
}