Continuous Same Game (2)

After repeated attempts, LL finds the greedy strategy is very awful in practice. Even there is no apparent evidence to proof it is better than a random one. So he has to drop this strategy and try to discover a better one.

There are 100 test cases. Each test case begins with two integers n,m ( 5<=n,m<=20 ), which is the size of the board. Then n lines follow, each contains m characters, indicating the color of the block. There are 5 colors, and each with equal probability.

For each test case, first output a single line containing the number of steps S. Then S lines follow, each contains two integers indicating the position of an arbitrary block of the group you want to remove.

For the i-th test case, let Ai is the total point of your scheme and Bi is the total point of the greedy strategy . Then I will calculate the average of (Ai/Bi), 1<=i<=100. You will get WA if your ouput is invalid or this average is less than 1.5.

5 5
35552
31154
33222
21134
12314

5
2 2
0 0
1 1
2 1
4 4

35552 35000 00000 00000 00000 00000
31154 31552 05552 00002 00002 00000
33222 33154 01154 01104 00004 00000
21134 21134 21134 21134 20034 20030
12314 12314 12314 12314 12314 12312

The total point is 6+12+12+12+6=48.