Time Limit: 15000/8000 MS (Java/Others)

Memory Limit: 102400/102400 K (Java/Others)

Doge, tired of being a popular image on internet, is considering moving to another city for a new way of life.

In his country there are N (2 ≤N≤ 1000) cities labeled 0 . . . N - 1. He is currently in city 0. Meanwhile, for each pair of cities, there exists a road connecting them, costing C_{i},_{j} (a positive integer) for traveling from city i to city j. Please note that C_{i},_{j} may not equal to C_{j},_{i} for any given i ≠ j.

Doge is carefully examining the cities: in fact he will divide cities (his current city 0 is**NOT** included) into M (2 ≤ M ≤ 10^{6}) categories as follow: If the minimal cost from his current city (labeled 0) to the city i is Di, city i belongs to category numbered **D**_{i} mod M.Doge wants to know the “minimal” category (a category with minimal number) which contains at least one city.

For example, for a country with 4 cities (labeled 0 . . . 3, note that city 0 is not considered), Doge wants to divide them into 3 categories. Suppose category 0 contains no city, category 1 contains city 2 and 3, while category 2 contains city 1, Doge consider category 1 as the minimal one.

Could you please help Doge solve this problem?

Note:

C_{i},_{j} is generated in the following way:

Given integers X_{0}, X_{1}, Y_{0}, Y_{1}, (1 ≤ X_{0}, X_{1}, Y_{0}, Y_{1}≤ 1234567), for k ≥ 2 we have

Xk = (12345 + X_{k-1} * 23456 + X_{k-2} * 34567 + X_{k-1} * X_{k-2} * 45678) mod 5837501

Yk = (56789 + Y_{k-1} * 67890 + Y_{k-2} * 78901 + Y_{k-1} * Y_{k-2} * 89012) mod 9860381

The for k ≥ 0 we have

Z_{k} = (X_{k} * 90123 + Y_{k} ) mod 8475871 + 1

Finally for 0 ≤ i, j ≤ N - 1 we have

C_{i},_{j} = Z_{i*n+j} for i ≠ j

C_{i},_{j} = 0 for i = j

In his country there are N (2 ≤N≤ 1000) cities labeled 0 . . . N - 1. He is currently in city 0. Meanwhile, for each pair of cities, there exists a road connecting them, costing C

Doge is carefully examining the cities: in fact he will divide cities (his current city 0 is

For example, for a country with 4 cities (labeled 0 . . . 3, note that city 0 is not considered), Doge wants to divide them into 3 categories. Suppose category 0 contains no city, category 1 contains city 2 and 3, while category 2 contains city 1, Doge consider category 1 as the minimal one.

Could you please help Doge solve this problem?

Note:

C

Given integers X

Xk = (12345 + X

Yk = (56789 + Y

The for k ≥ 0 we have

Z

Finally for 0 ≤ i, j ≤ N - 1 we have

C

C

There are several test cases. Please process till EOF.

For each test case, there is only one line containing 6 integers N,M,X_{0},X_{1},Y_{0},Y_{1}.See the description for more details.

For each test case, there is only one line containing 6 integers N,M,X

For each test case, output a single line containing a single integer: the number of minimal category.

1 10 For the first test case, we have 0 1 2 3 4 5 6 7 8 X 1 2 185180 788997 1483212 4659423 4123738 2178800 219267 Y 3 4 1633196 7845564 2071599 4562697 3523912 317737 1167849 Z 90127 180251 1620338 2064506 625135 5664774 5647950 8282552 4912390 the cost matrix C is 0 180251 1620338 2064506 0 5664774 5647950 8282552 0So the minimal cost from city 0 to city 1 is 180251, while the distance to city 2 is 1620338. Given M = 10, city 1 and city 2 belong to category 1 and 8 respectively. Since only category 1 and 8 contain at least one city, the minimal one of them, category 1, is the desired answer to Doge’s question.Hint

提交代码