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 C++14 (GCC 5.4.1) (judge verdict: Accepted)



#include "bits/stdc++.h"

#include <unordered_set>
#include <unordered_map>

//#include <iostream>//POJ
//#include <vector>
//#include <string>
//#include <iomanip>
//#include <math.h>
//#include <algorithm>
//#include <cstring>
using namespace std;
#define l0 (lint)0
#define setc cin.tie(0);ios::sync_with_stdio(0)
#define dd(n) cout<<fixed<<setprecision(n)
#define inp(x) (cin>>x,x)
#define ind(x) (cin>>x,--x)
//#define repi(i,a,b) for(int i=(a), i##_len=(b); i<i##_len; ++i)
#define repi(i,a,b) for(int i=(a);i<b;++i)
#define rep(i,n) repi(i,0,n)
#define repi_(i,a,b) for(int i=(a), i##_len=(b); i<=i##_len; ++i)
#define rep_(i,n) repi_(i,0,n)
#define repir(i,a,b) for(int i=(a)-1, i##_first=(b); i>=i##_first; --i)
#define repr(i,n) repir(i,n,0)
#define repir_(i,a,b) for(int i=(a), i##_first=(b); i>=i##_first; --i)
#define repr_(i,n) repir_(i,n,0)
#define prep(i,n) for(int i=0;i*i<n;++i)
#define prepi(i,a,b) for(int i=a;i*i<b;++i)
#define all(x) (x).begin(),(x).end()
#define vsort(v) sort((v).begin(), (v).end())
#define vcsort(v,c) sort((v).begin(),(v).end(),c)
#define vrev(v) reverse((v).begin(), (v).end())
#define fnd(v,x) ((v).find(x)!=string::npos)
#define mod(a,b) (a<0?a%b+abs(b):a%b)
#define cl(a,b) (a+b-1/b)

//template<typename T> T vpop(vector<T> *v) {
//	T d = v->back();
//	v->pop_back();
//	return d;
//}
//char vpop(string *v) {
//	char d = v->back();
//	v->pop_back();
//	return d;
//}
#define siz(v) ((int)(v).size())
#define ers(v, n) (v).erase((v).begin() + n)
#define cnt(v, n) count(all(v), n)
#define vmin(v) *min_element(v.begin(), v.end())
#define vmax(v) *max_element(v.begin(), v.end())
#define contain(q) !q.empty()
#define cont(q) !q.empty()
//#define qpop(q, a, b) a=q.back().first;b=q.back().second;q.pop()
//#define pqpop(p, a, b) a=q.top().first;b=q.top().second;q.pop()
#define el "\n"
#define sp " "
#define pi 3.14159265358979
#define co(x) cout<<x<<el
#define coc(c, a, b) co(((c)?a:b))
#define cim(x) {cin>>x;--x;}
#define cim2(a,b) {cin>>a>>b;--a;--b;}
#define cosp(x) cout<<(x)<<' '
#define YES(c) coc(c,"YES", "NO")
#define Yes(c) coc(c,"Yes", "No")
#define yes(c) coc(c,"yes", "no")
#define POSSIBLE(c) coc(c, "POSSIBLE", "IMPOSSIBLE")
#define Possible(c) coc(c, "Possible", "Impossible")
#define possible(c) coc(c, "possible", "impossible")
#define inf INT_MAX
#define linf INT64_MAX*0.99
#define wildcard(T) numeric_limits<T>::min()
#define noway() {co(-1);gc();return 0;}

#define pb push_back
#define pq priority_queue
#define np next_permutation

#ifdef tqktmp_2
#define gc() getchar(); getchar()
#else
#define gc() 1
#endif
const unsigned int bf0 = (1 << 0);
const unsigned int bf1 = (1 << 1);
const unsigned int bf2 = (1 << 2);
const unsigned int bf3 = (1 << 3);
const unsigned int bf4 = (1 << 4);
const unsigned int bf5 = (1 << 5);
const unsigned int bf6 = (1 << 6);
const unsigned int bf7 = (1 << 7);
const int _10j9p7 = 1000000007;

//#define lint long long
typedef long long lint;
typedef vector<int> IV; typedef vector<string> SV;
typedef vector<lint> LIV;
typedef vector<vector<int>> IVV;
typedef pair<int,int> P; typedef pair<lint,lint> LP;
typedef vector<P> PV; typedef vector<LP> LPV;

const int
dx8[8] ={0,1,1,1,0,-1,-1,-1},
dy8[8] ={1,1,0,-1,-1,-1,0,1},
dx9[9] ={0,1,1,1,0,0,-1,-1,-1},
dy9[9] ={1,1,0,-1,0,-1,-1,0,1},
dx5[5] ={0,1,0,0,-1},
dy5[5] ={1,0,0,-1,0},
dx4[4] ={0,1,0,-1},
dy4[4] ={1,0,-1,0};
//template<class... A> void cim_(A... args) {
//	for (A *i : initializer_list<A*>{ args... }) {
//		cin>>*i;--*i;
//	}
//	return;
//}

template<class T>inline bool maxi(T &a,const T &b){ if(a < b){ a = b; return 1; } return 0; }
template<class T>inline bool mini(T &a,const T &b){ if(b < a){ a = b; return 1; } return 0; }


#define MAX_N 1000
const int md = 1e9+7;

struct edge{
	int to,cost;
};
//graph--------------------------//
int n;					//size
vector<edge> g[MAX_N];	//edges
//-------------------------------//

struct dii{//for pfs
	int dist,	//distance
		i;		//vertex number
	bool operator>(const dii &r) const{
		return dist>r.dist;
	}bool operator<(const dii &r) const{
		return dist<r.dist;
	}
};
int d[MAX_N];			//distance
void pfs(int s){//start
	priority_queue<dii> q;
	fill(d,d+n,inf*0.9);d[s]=0;q.push({0,s});
	int cur;
	edge e;
	while(!q.empty()){
		dii p=q.top();q.pop();
		cur=p.i;
		if(d[cur]<p.dist)continue;
		rep(i,siz(g[cur])){
			e=g[cur][i];
			if(d[e.to]>d[cur]+e.cost){
				d[e.to]=d[cur]+e.cost;
				q.push({d[e.to],e.to});
			}
		}
	}
}

int main(){
	setc;

	lint h,w;cin>>h>>w;
	int c;n=10;
	rep(j,10){
		rep(i,10){
			cin>>c;
			g[j].pb({i,c});
		}
	}
	lint costs[10];
	rep(i,10){
		pfs(i);
		costs[i]=d[1];
	}
	lint x,ans=0;
	rep(j,h){
		rep(i,w){
			cin>>x;
			if(x!=-1)ans+=costs[x];
		}
	}co(ans);
	gc();
}