Problem:
Problem C: One-Dimensional Cellular Automaton

There is a one-dimensional cellular automaton consisting of N cells. Cells are numbered from 0 to N − 1.

Each cell has a state represented as a non-negative integer less than M. The states of cells evolve through discrete time steps. We denote the state of the i-th cell at time t as S(i, t). The state at time t + 1 is defined by the equation

S(i, t + 1) = (A × S(i − 1, t) + B × S(i, t) + C × S(i + 1, t)) mod M,         (1)

where A, B and C are non-negative integer constants. For i < 0 or N ≤ i, we define S(i, t) = 0.

Given an automaton definition and initial states of cells, your mission is to write a program that computes the states of the cells at a specified time T.

Input

The input is a sequence of datasets. Each dataset is formatted as follows.

N M A B C T

S(0, 0) S(1, 0) ... S(N − 1, 0)

The first line of a dataset consists of six integers, namely N, M, A, B, C and T. N is the number of cells. M is the modulus in the equation (1). A, B and C are coefficients in the equation (1). Finally, T is the time for which you should compute the states.

You may assume that 0 < N ≤ 50, 0 < M ≤ 1000, 0 ≤ A, B, C < M and 0 ≤ T ≤ 109.

The second line consists of N integers, each of which is non-negative and less than M. They represent the states of the cells at time zero.

A line containing six zeros indicates the end of the input.

Output

For each dataset, output a line that contains the states of the cells at time T. The format of the output is as follows.

S(0, T) S(1, T) ... S(N − 1, T)

Each state must be represented as an integer and the integers must be separated by a space.

Sample Input

5 4 1 3 2 0
0 1 2 0 1
5 7 1 3 2 1
0 1 2 0 1
5 13 1 3 2 11
0 1 2 0 1
5 5 2 0 1 100
0 1 2 0 1
6 6 0 2 3 1000
0 1 2 0 1 4
20 1000 0 2 3 1000000000
0 1 2 0 1 0 1 2 0 1 0 1 2 0 1 0 1 2 0 1
30 2 1 0 1 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
30 2 1 1 1 1000000000
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
30 5 2 3 1 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0

Output for the Sample Input

0 1 2 0 1
2 0 0 4 3
2 12 10 9 11
3 0 4 2 1
0 4 2 0 4 4
0 376 752 0 376 0 376 752 0 376 0 376 752 0 376 0 376 752 0 376
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 3 2 2 2 3 3 1 4 3 1 2 3 0 4 3 3 0 4 2 2 2 2 1 1 2 1 3 0

Solution in Kotlin (judge verdict: Accepted)

fun main(args: Array<String>?): Unit {
    while(true){
        val (n, m, a, b, c, t) = readLine()!!.trim().split(' ').map(String::toInt)
        if (n == 0) return
        val s = readLine()!!.trim().split(' ').map{it.toInt() % m}
        val matrix = Array(n){i -> IntArray(n){j -> if (i - 1 == j) a else if (i == j) b else if (i + 1 == j) c else 0}}
        println(
                times(matrix.exp(t, m), s.toIntArray(), m).joinToString(" ")
        )
    }
}
operator fun List<Int>.component6(): Int = this[5]

fun times(a: Array<IntArray>, b: Array<IntArray>, m: Int): Array<IntArray> {
    val result = Array(a.size){IntArray(b.first().size){0}}
    for (i in result.indices) {
        for (j in result[i].indices) {
            for (k in a[i].indices) {
                result[i][j] += a[i][k] * b[k][j]
            }
            result[i][j] %= m
        }
    }
    return result
}
fun times(a: Array<IntArray>, b: IntArray, m: Int): IntArray {
    return times(a, b.map{ intArrayOf(it)}.toTypedArray(), m).map{it.first()}.toIntArray()
}
fun double(a: Array<IntArray>, m: Int): Array<IntArray> {
    return times(a, a, m)
}
fun Array<IntArray>.exp(p: Int, m: Int): Array<IntArray> {
    return when(p){
        0 -> Array(this.size){i -> IntArray(this.size){j -> if (i == j) 1 else 0}}
        1 -> this
        else -> if (p % 2 == 0) double(this.exp(p / 2, m), m) else times(this, double(this.exp(p / 2, m), m), m)
    }
}