1">1×11×1 square plots aligned with north-south and east-west directions. Each plot can be uniquely described by integer Cartesian coordinates (x,y)(x,y) of its south-west corner.

Three friends, Alice, Bob, and Charlie are going to buy three distinct plots of land A,B,C

A,B,C in the forest. Initially, all plots in the forest (including the plots A,B,CA,B,C) are covered by trees. The friends want to visit each other, so they want to clean some of the plots from trees. After cleaning, one should be able to reach any of the plots A,B,CA,B,C from any other one of those by moving through adjacent cleared plots. Two plots are adjacent if they share a side.

For example, A=(0,0)A=(0,0), B=(1,1)B=(1,1), C=(2,2)C=(2,2). The minimal number of plots to be cleared is 55. One of the ways to do it is shown with the gray color.

Of course, the friends don't want to strain too much. Help them find out the smallest number of plots they need to clean from trees.

Input

The first line contains two integers xAxA and yAyA — coordinates of the plot AA (0≤xA,yA≤10000≤xA,yA≤1000). The following two lines describe coordinates (xB,yB)(xB,yB) and (xC,yC)(xC,yC) of plots BB and CC respectively in the same format (0≤xB,yB,xC,yC≤10000≤xB,yB,xC,yC≤1000). It is guaranteed that all three plots are distinct.

Output

On the first line print a single integer kk — the smallest number of plots needed to be cleaned from trees. The following kk lines should contain coordinates of all plots needed to be cleaned. All kk plots should be distinct. You can output the plots in any order.

If there are multiple solutions, print any of them.

Examples input Copy

0 0
1 1
2 2

output Copy

5
0 0
1 0
1 1
1 2
2 2

input Copy

0 0
2 0
1 1

output Copy

4
0 0
1 0
1 1
2 0

Note

The first example is shown on the picture in the legend.

The second example is illustrated with the following image:

 

 

題意概括：

給三個格子的座標，要求用最少的格子把這三個格子連起來，要求相鄰格子相連需要是要有公共邊。

 

解題思路：

所需要的總步數就是 X軸方向最大差值  加 Y軸方向最大差值 加 1.

輸出的方格：

先按 X 小 Y 大的優先順序對三個座標排序。

從第一個點出發到第二個點，採取先沿著 X 軸 方向走，後沿著 Y軸 方向走，同時記錄離第三個點的曼哈頓距離最近的一個轉折點。

從轉折點走到第三個點，採取先沿著Y軸方向走，後沿著 X軸方向走。

tip：如果擔心會走重複的格子，加個標記就可以了。

 

AC code：

 1 #include <queue>
 2 #include <cmath>
 3 #include <cstdio>
 4 #include <vector>
 5 #include <cstring>
 6 #include <iostream>
 7 #include <algorithm>
 8 #define LL long long
 9 using namespace std;
10 
11 const int INF = 0x3f3f3f3f;
12 const int MAXN = 1e3+10;
13 int N;
14 
15 struct date
16 {
17     int x, y;
18 }index[4];
19 
20 bool cmp(date a, date b)
21 {
22     if(a.x != b.x) return a.x < b.x;
23     else return a.y > b.y;
24 }
25 
26 bool mmp[MAXN][MAXN];
27 
28 int main()
29 {
30     date kk, nxt;
31     int maxx = 0, maxy = 0, minx = INF, miny = INF;
32     for(int i = 1; i <= 3; i++){
33         scanf("%d %d", &index[i].x, &index[i].y);
34         maxx = max(maxx, index[i].x);
35         maxy = max(maxy, index[i].y);
36         minx = min(minx, index[i].x);
37         miny = min(miny, index[i].y);
38     }
39     sort(index+1, index+4, cmp);
40     memset(mmp, 1, sizeof(mmp));
41     int ans = (maxx-minx)+(maxy-miny)+1;
42     printf("%d\n", ans);
43     printf("%d %d\n", index[1].x, index[1].y);
44     kk.x = index[1].x;
45     kk.y = index[1].y;
46     nxt.x = index[2].x;
47     nxt.y = index[2].y;
48     mmp[kk.x][kk.y] = false;
49 
50     int len = abs(index[2].x - index[1].x);
51     for(int i = 1; i <= len; i++){
52         kk.x++;
53         if(mmp[kk.x][kk.y]) printf("%d %d\n", kk.x, kk.y);
54         if(kk.x == index[3].x){
55                 nxt.x = kk.x;
56                 nxt.y = kk.y;
57         }
58         mmp[kk.x][kk.y] = false;
59     }
60 
61     len = abs(index[1].y - index[2].y);
62     for(int i = 1; i <= len; i++){
63         if(index[1].y < index[2].y) kk.y++;
64         else kk.y--;
65         if(mmp[kk.x][kk.y]) printf("%d %d\n", kk.x, kk.y);
66         if(kk.y == index[3].y){
67                 nxt.x = kk.x;
68                 nxt.y = kk.y;
69         }
70         mmp[kk.x][kk.y] = false;
71     }
72 
73    // printf("nxtx:%d nxty:%d\n", nxt.x, nxt.y);
74 
75     len = abs(index[3].y - nxt.y);
76     for(int i = 1; i <= len; i++){
77         if(nxt.y < index[3].y) nxt.y++;
78         else nxt.y--;
79         if(mmp[nxt.x][nxt.y]) printf("%d %d\n", nxt.x, nxt.y);
80         mmp[nxt.x][nxt.y] = false;
81     }
82 
83     len = abs(index[3].x - nxt.x);
84     for(int i = 1; i <= len; i++){
85         nxt.x++;
86         if(mmp[nxt.x][nxt.y]) printf("%d %d\n", nxt.x, nxt.y);
87         mmp[nxt.x][nxt.y] = false;
88     }
89 
90     return 0;
91 }