#include <string>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;

const int DX[] = {0, 1, 0, -1};
const int DY[] = {1, 0, -1, 0};
const char DIR[] = {'R', 'D', 'L', 'U'};

int N, K;
vector<vector<int>> grid;

int main() {
    auto current_time = high_resolution_clock::now();
    cin >> N >> K;
    grid = vector<vector<int>>(N, vector<int>(N));
    for (int i = 0; i < N; i++) {
        string s;
        cin >> s;
        for (int j = 0; j < N; j++) {
            grid[i][j] = s[j] - '0';
        }
    }
    vector<vector<double>> score(N, vector<double>(N, 0));
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (grid[i][j] == 0) continue;
            deque<pair<int, int>> q;
            q.push_back({i, j});
            vector<vector<bool>> visited(N, vector<bool>(N, false));
            vector<vector<double>> now_score(N, vector<double>(N, 0));
            now_score[i][j] = 0.1 * grid[i][j];
            while (!q.empty()) {
                auto [x, y] = q.front();
                q.pop_front();
                if (visited[x][y]) continue;
                visited[x][y] = true;
                double p = 1.0;
                for (int d = 0; d < 4; d++) {
                    int nx = x + DX[d];
                    int ny = y + DY[d];
                    if (nx < 0 || nx >= N || ny < 0 || ny >= N) continue;
                    if (grid[nx][ny] == 0) continue;
                    p *= (1.0 - now_score[nx][ny]);
                    q.push_back({nx, ny});
                }
                now_score[x][y] += (1.0 - p) * 0.1 * grid[x][y];
            }
            for (int x = 0; x < N; x++) {
                for (int y = 0; y < N; y++) {
                    score[x][y] += now_score[x][y];
                }
            }
        }
    }
    vector<pair<double, int>> ans;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            ans.push_back({score[i][j], i * N + j});
        }
    }
    sort(ans.rbegin(), ans.rend());
    cout << K << endl;
    for (int i = 0; i < K; i++) {
        cout << ans[i].second / N << " " << ans[i].second % N << endl;
    }
}