Problem: Smart Beaver became interested in drawing. He draws suns. However, at some point, Smart Beaver realized that simply drawing suns is boring. So he decided to design a program that will process his drawings. You are given a picture drawn by the beaver. It will have two colors: one for the background and one for the suns in the image. Your task will be to count the number of suns in the image and for each of them to count the number of rays.

Sun is arbitrarily rotated ellipse with rays. Ray is a segment which connects point on boundary of the ellipse with some point outside ellipse.

<image> An image where all suns are circles.  <image> An image where all suns are ellipses, their axes are parallel to the coordinate axes.  <image> An image where all suns are rotated ellipses. 

It is guaranteed that: 

  * No two suns have common points. 
  * The rays’ width is 3 pixels. 
  * The lengths of the ellipsis suns’ axes will lie between 40 and 200 pixels. 
  * No two rays intersect. 
  * The lengths of all rays will lie between 10 and 30 pixels. 

Input

The first line contains two integers h and w — the height and width of the image (1 ≤ h, w ≤ 1600). Next h lines will contain w space-separated integers each. They describe Smart Beaver’s picture. Each number equals either a 0 (the image background), or a 1 (the sun color).

The input limits for scoring 30 points are (subproblem F1): 

  * All suns on the image are circles. 



The input limits for scoring 70 points are (subproblems F1+F2): 

  * All suns on the image are ellipses with axes parallel to the coordinate axes. 



The input limits for scoring 100 points are (subproblems F1+F2+F3):

  * All suns on the image are ellipses, they can be arbitrarily rotated. 

Output

The first line must contain a single number k — the number of suns on the beaver’s image. The second line must contain exactly k space-separated integers, corresponding to the number of rays on each sun. The numbers of the second line must be sorted in the increasing order.

Examples

Note

For each complexity level you are suggested a sample in the initial data. You can download the samples at http://www.abbyy.ru/sun.zip.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:20000000")
using namespace std;
template <typename T1, typename T2, typename T3>
struct triple {
  T1 a;
  T2 b;
  T3 c;
  triple(){};
  triple(T1 _a, T2 _b, T3 _c) : a(_a), b(_b), c(_c) {}
};
template <typename T1, typename T2, typename T3>
bool operator<(const triple<T1, T2, T3> &t1, const triple<T1, T2, T3> &t2) {
  if (t1.a != t2.a)
    return t1.a < t2.a;
  else
    return t1.b < t2.b;
}
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::pair<T1, T2> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::vector<T> &v) {
  bool first = true;
  os << "[";
  for (unsigned int i = 0; i < v.size(); i++) {
    if (!first) os << ", ";
    os << v[i];
    first = false;
  }
  return os << "]";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::set<T> &v) {
  bool first = true;
  os << "[";
  for (typename std::set<T>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::map<T1, T2> &v) {
  bool first = true;
  os << "[";
  for (typename std::map<T1, T2>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T, typename T2>
void printarray(T a[], T2 sz, T2 beg = 0) {
  for (T2 i = beg; i < sz; i++) cout << a[i] << " ";
}
inline long long binpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res *= x;
    x *= x;
    n >>= 1;
  }
  return res;
}
inline long long powmod(long long x, long long n, long long _mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x) % _mod;
    x = (x * x) % _mod;
    n >>= 1;
  }
  return res;
}
inline long long mulmod(long long x, long long n, long long _mod) {
  long long res = 0;
  while (n) {
    if (n & 1) res = (res + x) % _mod;
    x = (x + x) % _mod;
    n >>= 1;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  long long t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline int gcd(int a, int b) {
  int t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline long long lcm(int a, int b) { return a / gcd(a, b) * (long long)b; }
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long gcd(long long a, long long b, long long c) {
  return gcd(gcd(a, b), c);
}
inline int gcd(int a, int b, int c) { return gcd(gcd(a, b), c); }
inline long long lcm(long long a, long long b, long long c) {
  return lcm(lcm(a, b), c);
}
inline long long lcm(int a, int b, int c) {
  return lcm(lcm(a, b), (long long)c);
}
inline long long max(long long a, long long b) { return (a > b) ? a : b; }
inline int max(int a, int b) { return (a > b) ? a : b; }
inline double max(double a, double b) { return (a > b) ? a : b; }
inline long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
inline int max(int a, int b, int c) { return max(a, max(b, c)); }
inline double max(double a, double b, double c) { return max(a, max(b, c)); }
inline long long min(long long a, long long b) { return (a < b) ? a : b; }
inline int min(int a, int b) { return (a < b) ? a : b; }
inline double min(double a, double b) { return (a < b) ? a : b; }
inline long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
inline int min(int a, int b, int c) { return min(a, min(b, c)); }
inline double min(double a, double b, double c) { return min(a, min(b, c)); }
inline void getar(int *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%d", a + ii);
  }
}
inline void getar(long long *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%I64", a + ii);
  }
}
bool a[1621][1621];
bool used[1621][1621];
bool used2[1621][1621];
bool used3[1621][1621];
const int maxl = 15;
const int maxs = 10;
int w, h;
int k = 0;
int cur = 0;
int basei, basej;
triple<int, int, int> q[10000];
int qs = 0;
void go2(int i, int j) {
  if (((abs(i - basei) + abs(j - basej)) >= maxl) || (!a[i][j]) || used2[i][j])
    return;
  used2[i][j] = true;
  qs = 0;
  q[qs].a = i;
  q[qs].b = j;
  q[qs++].c = 0;
  int qc = 0;
  while (qc < qs) {
    int curi = q[qc].a;
    int curj = q[qc].b;
    int curl = q[qc++].c;
    if (curl == maxl) break;
    for (int t = -1; t <= 1; t++)
      for (int w = -1; w <= 1; w++) {
        if (!used2[curi + t][curj + w] && a[curi + t][curj + w]) {
          used2[curi + t][curj + w] = true;
          q[qs].a = curi + t;
          q[qs].b = curj + w;
          q[qs++].c = curl + 1;
        }
      }
  }
}
vector<pair<int, int>> v3;
bool check(int i, int j) {
  int s = 0;
  qs = 0;
  q[qs].a = i;
  q[qs].b = j;
  q[qs++].c = 0;
  used3[i][j] = true;
  int qc = 0;
  v3.clear();
  while (qc < qs) {
    int curi = q[qc].a;
    int curj = q[qc].b;
    int curl = q[qc++].c;
    if (max(abs(basei - curi), abs(basej - curj)) > 2) continue;
    if (a[curi][curj]) s++;
    for (int t = -1; t <= 1; t++)
      for (int w = -1; w <= 1; w++) {
        if (!used3[curi + t][curj + w]) {
          used3[curi + t][curj + w] = true;
          v3.push_back(make_pair(curi + t, curj + w));
          q[qs].a = curi + t;
          q[qs].b = curj + w;
          q[qs++].c = curl + 1;
        }
      }
  }
  for (int i = 0; i < v3.size(); i++) {
    used3[v3[i].first][v3[i].second] = false;
  }
  return (s < maxs);
}
void go(int i, int j) {
  if ((!a[i][j]) || used[i][j]) return;
  used[i][j] = true;
  go(i - 1, j);
  go(i + 1, j);
  go(i, j - 1);
  go(i, j + 1);
  if (used2[i][j]) return;
  int t = 0;
  if (a[i - 1][j] == 0) t++;
  if (a[i][j - 1] == 0) t++;
  if (a[i + 1][j] == 0) t++;
  if (a[i][j + 1] == 0) t++;
  if (t >= 2) {
    basei = i;
    basej = j;
    if (check(i, j)) {
      cur++;
      go2(i, j);
    }
  }
}
vector<int> ans;
int main() {
  scanf("%d", &w);
  scanf("%d", &h);
  int x;
  for (int i = 5; i < w + 5; i++) {
    for (int j = 5; j < h + 5; j++) {
      scanf("%d", &x);
      a[i][j] = x;
      used[i][j] = used2[i][j] = false;
    }
  }
  for (int i = 5; i < w + 5; i++) {
    for (int j = 5; j < h + 5; j++) {
      if ((!used[i][j]) && a[i][j]) {
        k++;
        cur = 0;
        go(i, j);
        ans.push_back(cur);
      }
    }
  }
  cout << k << endl;
  sort(ans.begin(), ans.end());
  for (int i = 0; i < k; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}