Problem: You are given a rectangular grid of lattice points from (0, 0) to (n, m) inclusive. You have to choose exactly 4 different points to build a polyline possibly with self-intersections and self-touching. This polyline should be as long as possible.

A polyline defined by points p1, p2, p3, p4 consists of the line segments p1 p2, p2 p3, p3 p4, and its length is the sum of the lengths of the individual line segments.

Input

The only line of the input contains two integers n and m (0 ≤ n, m ≤ 1000). It is guaranteed that grid contains at least 4 different points.

Output

Print 4 lines with two integers per line separated by space — coordinates of points p1, p2, p3, p4 in order which represent the longest possible polyline.

Judge program compares your answer and jury's answer with 10 - 6 precision.

Examples

Input

1 1


Output

1 1
0 0
1 0
0 1


Input

0 10


Output

0 1
0 10
0 0
0 9

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void output_point(pair<int, int> p, bool swapped) {
  if (swapped)
    cout << p.second << " " << p.first << endl;
  else
    cout << p.first << " " << p.second << endl;
}
inline double MY_SQR(int x) { return ((double)x) * x; }
double dist(const pair<int, int> &p1, const pair<int, int> &p2) {
  return sqrt(MY_SQR(p1.first - p2.first) + MY_SQR(p1.second - p2.second));
}
int main(int argc, char *argv[]) {
  int n, m;
  cin >> n >> m;
  bool swapped = false;
  if (n > m) {
    swap(n, m);
    swapped = true;
  }
  vector<pair<int, int> > used_v;
  used_v.push_back(pair<int, int>(0, 0));
  used_v.push_back(pair<int, int>(n, m));
  vector<pair<int, int> > all_near_borders;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++)
      if (i == 0 || i == n || j == 0 || j == m)
        all_near_borders.push_back(pair<int, int>(i, j));
  all_near_borders.pop_back();
  all_near_borders.erase(all_near_borders.begin());
  double found_max = 0;
  vector<pair<int, int> > used_v_best;
  for (size_t k2 = 0; k2 < all_near_borders.size(); k2++) {
    for (int k3 = (int)all_near_borders.size() - 1; k3 >= 0; k3--)
      if (k3 != k2) {
        double curr_dist_1 = dist(used_v[1], all_near_borders[k2]) +
                             dist(all_near_borders[k2], all_near_borders[k3]);
        if (curr_dist_1 > found_max) {
          found_max = curr_dist_1;
          used_v_best = used_v;
          used_v_best.push_back(all_near_borders[k2]);
          used_v_best.push_back(all_near_borders[k3]);
        }
        double curr_dist_2 = dist(all_near_borders[k2], used_v[0]) +
                             dist(used_v[1], all_near_borders[k3]);
        if (curr_dist_2 > found_max) {
          found_max = curr_dist_2;
          used_v_best = used_v;
          used_v_best.insert(used_v_best.begin(), all_near_borders[k2]);
          used_v_best.push_back(all_near_borders[k3]);
        }
      }
  }
  for (int i = 0; i < 4; i++) output_point(used_v_best[i], swapped);
  return 0;
}