Problem:
Score : 400 points

Problem Statement

There is a grid of square cells with H horizontal rows and W vertical columns. The cell at the i-th row and the j-th column will be denoted as Cell (i, j).

In Cell (i, j), a_{ij} coins are placed.

You can perform the following operation any number of times:

Operation: Choose a cell that was not chosen before and contains one or more coins, then move one of those coins to a vertically or horizontally adjacent cell.

Maximize the number of cells containing an even number of coins.

Constraints

All values in input are integers.

1 \leq H, W \leq 500

0 \leq a_{ij} \leq 9

Input

Input is given from Standard Input in the following format:

H W
a_{11} a_{12} ... a_{1W}
a_{21} a_{22} ... a_{2W}
:
a_{H1} a_{H2} ... a_{HW}

Output

Print a sequence of operations that maximizes the number of cells containing an even number of coins, in the following format:

N
y_1 x_1 y_1' x_1'
y_2 x_2 y_2' x_2'
:
y_N x_N y_N' x_N'

That is, in the first line, print an integer N between 0 and H \times W (inclusive), representing the number of operations.

In the (i+1)-th line (1 \leq i \leq N), print four integers y_i, x_i, y_i' and x_i' (1 \leq y_i, y_i' \leq H and 1 \leq x_i, x_i' \leq W), representing the i-th operation. These four integers represents the operation of moving one of the coins placed in Cell (y_i, x_i) to a vertically or horizontally adjacent cell, (y_i', x_i').

Note that if the specified operation violates the specification in the problem statement or the output format is invalid, it will result in Wrong Answer.

Sample Input 1

2 3
1 2 3
0 1 1

Sample Output 1

3
2 2 2 3
1 1 1 2
1 3 1 2

Every cell contains an even number of coins after the following sequence of operations:

Move the coin in Cell (2, 2) to Cell (2, 3).

Move the coin in Cell (1, 1) to Cell (1, 2).

Move one of the coins in Cell (1, 3) to Cell (1, 2).

Sample Input 2

3 2
1 0
2 1
1 0

Sample Output 2

3
1 1 1 2
1 2 2 2
3 1 3 2

Sample Input 3

1 5
9 9 9 9 9

Sample Output 3

2
1 1 1 2
1 3 1 4

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define ll int64_t
#define _GLIBCXX_DEBUG
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
//拡張ユークリッドの互除法
template< typename T >
T extgcd(T a, T b, T &x, T &y) {
  T d = a;
  if(b != 0) {
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
  return d;
}

//約数列挙(√N)
vector< int64_t > divisor(int64_t n) {
  vector< int64_t > ret;
  for(int64_t i = 1; i * i <= n; i++) {
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}
//素因数分解(√N)
map< int64_t, int > prime_factor(int64_t n) {
  map< int64_t, int > ret;
  for(int64_t i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}
//素数テーブル(NloglogN)
vector< bool > prime_table(int n) {
  vector< bool > prime(n + 1, true);
  if(n >= 0) prime[0] = false;
  if(n >= 1) prime[1] = false;
  for(int i = 2; i * i <= n; i++) {
    if(!prime[i]) continue;
    for(int j = i + i; j <= n; j += i) {
      prime[j] = false;
    }
  }
  return prime;
}
//二項係数(K)
template< typename T >
T binomial(int64_t N, int64_t K) {
  if(K < 0 || N < K) return 0;
  T ret = 1;
  for(T i = 1; i <= K; ++i) {
    ret *= N--;
    ret /= i;
  }
  return ret;
}
//二項係数テーブル(N^2)
template< typename T >
vector< vector< T > > binomial_table(int N) {
  vector< vector< T > > mat(N + 1, vector< T >(N + 1));
  for(int i = 0; i <= N; i++) {
    for(int j = 0; j <= i; j++) {
      if(j == 0 || j == i) mat[i][j] = 1;
      else mat[i][j] = mat[i - 1][j - 1] + mat[i - 1][j];
    }
  }
  return mat;
}
//Gragh template
template< typename T >
struct edge {
  int src, to;
  T cost;

  edge(int to, T cost) : src(-1), to(to), cost(cost) {}

  edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;
//unionfind
struct UnionFind {
  vector< int > data;
 
  UnionFind(int sz) {
    data.assign(sz, -1);
  }
 
  bool unite(int x, int y) {
    x = find(x), y = find(y);
    if(x == y) return (false);
    if(data[x] > data[y]) swap(x, y);
    data[x] += data[y];
    data[y] = x;
    return (true);
  }
 
  int find(int k) {
    if(data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }
 
  int size(int k) {
    return (-data[find(k)]);
  }
};
//DIJKSTRA(ElogV)最短路
template< typename T >
vector< T > dijkstra(WeightedGraph< T > &g, int s) {
  const auto INF = numeric_limits< T >::max();
  vector< T > dist(g.size(), INF);

  using Pi = pair< T, int >;
  priority_queue< Pi, vector< Pi >, greater< Pi > > que;
  dist[s] = 0;
  que.emplace(dist[s], s);
  while(!que.empty()) {
    T cost;
    int idx;
    tie(cost, idx) = que.top();
    que.pop();
    if(dist[idx] < cost) continue;
    for(auto &e : g[idx]) {
      auto next_cost = cost + e.cost;
      if(dist[e.to] <= next_cost) continue;
      dist[e.to] = next_cost;
      que.emplace(dist[e.to], e.to);
    }
  }
  return dist;
}
//prim最小全域木
template< typename T >
T prim(WeightedGraph< T > &g) {
  using Pi = pair< T, int >;
 
  T total = 0;
  vector< bool > used(g.size(), false);
  priority_queue< Pi, vector< Pi >, greater< Pi > > que;
  que.emplace(0, 0);
  while(!que.empty()) {
    auto p = que.top();
    que.pop();
    if(used[p.second]) continue;
    used[p.second] = true;
    total += p.first;
    for(auto &e : g[p.second]) {
      que.emplace(e.cost, e.to);
    }
  }
  return total;
}
//bellman_ford(VE)単一始点最短路
template< typename T >
vector< T > bellman_ford(Edges< T > &edges, int V, int s) {
  const auto INF = numeric_limits< T >::max();
  vector< T > dist(V, INF);
  dist[s] = 0;
  for(int i = 0; i < V - 1; i++) {
    for(auto &e : edges) {
      if(dist[e.src] == INF) continue;
      dist[e.to] = min(dist[e.to], dist[e.src] + e.cost);
    }
  }
  for(auto &e : edges) {
    if(dist[e.src] == INF) continue;
    if(dist[e.src] + e.cost < dist[e.to]) return vector< T >();
  }
  return dist;
}


//memo cout << fixed << setprecision(桁数);//
int main(){
  int h,w;
  cin >> h >> w;
  int n=0;
  vector<vector<int>> a(h,vector<int>(w));
  rep(i,h)
    rep(j,w)cin >> a.at(i).at(j);
  vector<int> e,b,c,d;
  rep(i,h){
    rep(j,w-1){
      if(a.at(i).at(j)%2==1){
    	e.push_back(i+1);
        b.push_back(j+1);
        c.push_back(i+1);
        d.push_back(j+2);
        n++;
        a.at(i).at(j)--;
        a.at(i).at(j+1)++;
      }
    }
  }
  rep(i,h-1){
    if(a.at(i).at(w-1)%2==1){
      e.push_back(i+1);
      b.push_back(w);
      c.push_back(i+2);
      d.push_back(w);      
      n++;
      a.at(i+1).at(w-1)++;
    }
  }
  cout << n << endl;
  rep(i,e.size())
    cout << e.at(i) << " "<<b.at(i)<<" "<<c.at(i)<<" "<<d.at(i) << endl;
}