#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<int>> count(N, vector<int>(N, 0));
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (grid[i][j] == 0) continue;
            for (int t = 0; t < 16; t++) {
                deque<pair<int, int>> q;
                q.push_back({i, j});
                vector<vector<bool>> visited(N, vector<bool>(N, false));
                visited[i][j] = true;
                while (!q.empty()) {
                    auto [x, y] = q.front();
                    q.pop_front();
                    if (rand() % 10 <= grid[x][y] - 1) {
                        count[x][y]++;
                    } else {
                        continue;
                    }
                    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 (visited[nx][ny]) continue;
                        if (grid[nx][ny] == 0) continue;
                        visited[nx][ny] = true;
                        q.push_back({nx, ny});
                    }
                }
            }
        }
    }
    vector<pair<int, int>> ans;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            ans.push_back({count[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;
    }
}