Codeforces887C-Solution for Cube

C. Solution for Cube

time limit per test

memory limit per test

input

output

During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2.

It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction.

To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above.

Cube is called solved if for each face of cube all squares on it has the same color.

​​https://en.wikipedia.org/wiki/Rubik's_Cube​​

Input

ai (1 ≤ ai), where ai denotes color of i-th square. There are exactly 4 occurrences of all colors in this sequence.

Output

YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise.

Examples

input

2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4

output

NO

input

5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3

output

YES

Note

In first test case cube looks like this:

In second test case cube looks like this:

It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16.

模拟。

Code:

var
  i,j,sum:longint;flag:boolean;
  a:array[1..6,1..4] of longint;
  f:array[1..6] of boolean;
begin
  for i:=1 to 6 do
  begin
    for j:=1 to 4 do read(a[i][j]);
    if (a[i][1]=a[i][2])and(a[i][1]=a[i][3])and(a[i][1]=a[i][4]) then
    begin
      f[i]:=true;
      inc(sum);
    end;
  end;
  if sum<>2 then writeln('NO') else
  begin
    if (f[1])and(f[3]) then
    begin
      flag:=true;
      for i:=1 to 6 do
        if (a[i][1]<>a[i][2])or(a[i][3]<>a[i][4]) then
          flag:=false;
      if flag then
      begin
        if ((a[4][1]=a[2][3])and(a[2][1]=a[5][3])and(a[5][1]=a[6][3])and(a[6][1]=a[4][3]))or
           ((a[4][3]=a[2][1])and(a[2][3]=a[5][1])and(a[5][3]=a[6][1])and(a[6][3]=a[4][1])) then
           writeln('YES') else writeln('NO');
      end else writeln('NO');
    end;
    if (f[4])and(f[5]) then
    begin
      flag:=true;
      for i:=1 to 6 do
        if (a[i][1]<>a[i][3])or(a[i][2]<>a[i][4]) then
          flag:=false;
      if flag then
      begin
        if ((a[1][1]=a[2][2])and(a[2][1]=a[3][2])and(a[3][1]=a[6][1])and(a[6][2]=a[1][2]))or
           ((a[1][2]=a[2][1])and(a[2][2]=a[3][1])and(a[3][2]=a[6][2])and(a[6][1]=a[1][1])) then
           writeln('YES') else writeln('NO');
      end else writeln('NO');
    end;
    if (f[2])and(f[6]) then
    begin
      flag:=true;
      if (a[1][1]<>a[1][2])or(a[1][3]<>a[1][4])or
         (a[3][1]<>a[3][2])or(a[3][3]<>a[3][4])or
         (a[4][1]<>a[4][3])or(a[4][2]<>a[4][4])or
         (a[5][1]<>a[5][3])or(a[5][2]<>a[5][4]) then flag:=false;
      if flag then
      begin
        if ((a[1][1]=a[5][1])and(a[5][2]=a[3][1])and(a[3][3]=a[4][2])and(a[4][1]=a[1][3]))or
           ((a[1][3]=a[5][2])and(a[5][1]=a[3][3])and(a[3][1]=a[4][1])and(a[4][2]=a[1][1])) then
           writeln('YES') else writeln('NO');
      end else writeln('NO');
    end;
  end;
end.