Problem:
Score : 400 points

Problem Statement

Joisino the magical girl has decided to turn every single digit that exists on this world into 1.

Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points).

She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive).

You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows:

If A_{i,j}≠-1, the square contains a digit A_{i,j}.

If A_{i,j}=-1, the square does not contain a digit.

Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.

Constraints

1≤H,W≤200

1≤c_{i,j}≤10^3 (i≠j)

c_{i,j}=0 (i=j)

-1≤A_{i,j}≤9

All input values are integers.

There is at least one digit on the wall.

Input

Input is given from Standard Input in the following format:

H W
c_{0,0} ... c_{0,9}
:
c_{9,0} ... c_{9,9}
A_{1,1} ... A_{1,W}
:
A_{H,1} ... A_{H,W}

Output

Print the minimum total amount of MP required to turn every digit on the wall into 1 in the end.

Sample Input 1

2 4
0 9 9 9 9 9 9 9 9 9
9 0 9 9 9 9 9 9 9 9
9 9 0 9 9 9 9 9 9 9
9 9 9 0 9 9 9 9 9 9
9 9 9 9 0 9 9 9 9 2
9 9 9 9 9 0 9 9 9 9
9 9 9 9 9 9 0 9 9 9
9 9 9 9 9 9 9 0 9 9
9 9 9 9 2 9 9 9 0 9
9 2 9 9 9 9 9 9 9 0
-1 -1 -1 -1
8 1 1 8

Sample Output 1

12

To turn a single 8 into 1, it is optimal to first turn 8 into 4, then turn 4 into 9, and finally turn 9 into 1, costing 6 MP.

The wall contains two 8s, so the minimum total MP required is 6×2=12.

Sample Input 2

5 5
0 999 999 999 999 999 999 999 999 999
999 0 999 999 999 999 999 999 999 999
999 999 0 999 999 999 999 999 999 999
999 999 999 0 999 999 999 999 999 999
999 999 999 999 0 999 999 999 999 999
999 999 999 999 999 0 999 999 999 999
999 999 999 999 999 999 0 999 999 999
999 999 999 999 999 999 999 0 999 999
999 999 999 999 999 999 999 999 0 999
999 999 999 999 999 999 999 999 999 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1

Sample Output 2

0

Note that she may not need to change any digit.

Sample Input 3

3 5
0 4 3 6 2 7 2 5 3 3
4 0 5 3 7 5 3 7 2 7
5 7 0 7 2 9 3 2 9 1
3 6 2 0 2 4 6 4 2 3
3 5 7 4 0 6 9 7 6 7
9 8 5 2 2 0 4 7 6 5
5 4 6 3 2 3 0 5 4 3
3 6 2 3 4 2 4 0 8 9
4 6 5 4 3 5 3 2 0 8
2 1 3 4 5 7 8 6 4 0
3 5 2 6 1
2 5 3 2 1
6 9 2 5 6

Sample Output 3

47

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

package main

import (
	"fmt"
)

func main() {
	var H, W int
	fmt.Scan(&H, &W)
	c := make([][]int, 10)
	for i := 0; i < 10; i++ {
		c[i] = make([]int, 10)
		for j := 0; j < 10; j++ {
			fmt.Scan(&c[i][j])
		}
	}

	minCosts := make([]int, 10)
	for i := 0; i < 10; i++ {
		minCosts[i] = solve(c, i)
	}

	ans := 0
	for i := 0; i < H; i++ {
		for j := 0; j < W; j++ {
			A := 0
			fmt.Scan(&A)
			if A == -1 || A == 1 {
				continue
			}
			ans += minCosts[A]
		}
	}

	fmt.Println(ans)
}

/** Return minimum cost from start to 1 */
func solve(c [][]int, start int) int {
	if start == 1 {
		return 0
	}
	const Big = 999999

	unfixed := make([]int, 10)
	fixed := make([]int, 10)
	for i := 0; i < 10; i++ {
		fixed[i] = Big
		unfixed[i] = Big
	}
	cur := start
	fixed[cur] = 0
	unfixed[cur] = 0

	// iterate 9 times
	for i := 0; i < 9; i++ {
		// calc unfixed costs.
		for j := 0; j < 10; j++ {
			// cur -> j
			if fixed[j] != Big {
				continue
			}
			x := fixed[cur] + c[cur][j]
			if unfixed[j] > x {
				unfixed[j] = x
			}
		}
		// find minimum unfixed costs.
		muc := Big
		for j := 0; j < 10; j++ {
			if fixed[j] == Big && unfixed[j] < muc {
				muc = unfixed[j]
				cur = j
			}
		}
		fixed[cur] = unfixed[cur]
	}
	return fixed[1]
}