Problem:
Score : 100 points

Problem Statement

There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.

For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j.
If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.

Find the number of different directed paths of length K in G, modulo 10^9 + 7.
We will also count a path that traverses the same edge multiple times.

Constraints

All values in input are integers.

1 \leq N \leq 50

1 \leq K \leq 10^{18}

a_{i, j} is 0 or 1.

a_{i, i} = 0

Input

Input is given from Standard Input in the following format:

N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}

Output

Print the number of different directed paths of length K in G, modulo 10^9 + 7.

Sample Input 1

4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0

Sample Output 1

6

G is drawn in the figure below:

There are six directed paths of length 2:

1 → 2 → 3

1 → 2 → 4

2 → 3 → 4

2 → 4 → 1

3 → 4 → 1

4 → 1 → 2

Sample Input 2

3 3
0 1 0
1 0 1
0 0 0

Sample Output 2

3

G is drawn in the figure below:

There are three directed paths of length 3:

1 → 2 → 1 → 2

2 → 1 → 2 → 1

2 → 1 → 2 → 3

Sample Input 3

6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0

Sample Output 3

1

G is drawn in the figure below:

There is one directed path of length 2:

4 → 5 → 6

Sample Input 4

1 1
0

Sample Output 4

0

Sample Input 5

10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0

Sample Output 5

957538352

Be sure to print the count modulo 10^9 + 7.

Solution in Go (1.6) (judge verdict: Accepted)

package main

import (
	"fmt"
	"bufio"
	"os"
	"strings"
	"strconv"
	"math"
)

var INF int32 = math.MaxInt32
var INF64 int64 = math.MaxInt64

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func max(a, b int) int {
	if a < b {
		return b
	}
	return a
}

func min64(a, b int64) int64 {
	if a < b {
		return a
	}
	return b
}

func max64(a, b int64) int64 {
	if a < b {
		return b
	}
	return a
}

func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}

func readLineNums(bf *bufio.Reader) []int {
	line, _ := bf.ReadString('\n')
	line = strings.Trim(line, "\n")
	fields := strings.Split(line, " ")
	nums := []int{}
	for _, field := range fields {
		num, _ := strconv.Atoi(field)
		nums = append(nums, num)
	}
	return nums
}

func readLineNum64s(bf *bufio.Reader) []int64 {
	line, _ := bf.ReadString('\n')
	line = strings.Trim(line, "\n")
	fields := strings.Split(line, " ")
	nums := []int64{}
	for _, field := range fields {
		num, _ := strconv.ParseInt(field, 10, 64)
		nums = append(nums, num)
	}
	return nums
}

func readLineFloats(bf *bufio.Reader) []float64 {
	line, _ := bf.ReadString('\n')
	line = strings.Trim(line, "\n")
	fields := strings.Split(line, " ")
	nums := []float64{}
	for _, field := range fields {
		num, _ := strconv.ParseFloat(field, 64)
		nums = append(nums, num)
	}
	return nums
}

func readLineStrs(bf *bufio.Reader) []string {
	line, _ := bf.ReadString('\n')
	line = strings.Trim(line, "\n")
	return strings.Split(line, " ")
}

func main() {
	bf := bufio.NewReader(os.Stdin)
	nk := readLineNum64s(bf)
	N, K := nk[0], nk[1]

	A := make([][]int64, N)
	for i := range A {
		A[i] = readLineNum64s(bf)
	}

	dp := make([][]int64, N) // dp[i][j]: i到j经过k步到达的方案数量
	t := make([][]int64, N)
	for i := range dp {
		dp[i] = make([]int64, N)
		t[i] = make([]int64, N)
		dp[i][i] = int64(1) // i到i经过0步就可以到达的方案数
	}

	var m int64 = 1000000007
	mul := func(a [][]int64, b [][]int64, c [][]int64) {
		for i:=0; i<int(N); i++ {
			for j:=0; j<int(N); j++ {
				c[i][j] = 0
				for k:=0; k<int(N); k++ {
					c[i][j] += a[i][k]*b[k][j]
					c[i][j] %= m
				}
			}
		}
	}

	for K > 0 {
		if K%2 == 1 {
			mul(dp, A, t) // 结果临时保存在ndp
			dp, t = t, dp
		}
		mul(A, A, t)
		A, t = t, A
		K /= 2
	}

	var ans int64
	for i:=0; i<int(N); i++ {
		for j:=0; j<int(N); j++ {
			ans += dp[i][j]
			if ans >= m {
				ans -= m
			}
		}
	}

	fmt.Println(ans)
}