Many students live in a dormitory. A dormitory is a whole new world of funny amusements and possibilities but it does have its drawbacks. 

There is only one shower and there are multiple students who wish to have a shower in the morning. That's why every morning there is a line of five people in front of the dormitory shower door. As soon as the shower opens, the first person from the line enters the shower. After a while the first person leaves the shower and the next person enters the shower. The process continues until everybody in the line has a shower.

Having a shower takes some time, so the students in the line talk as they wait. At each moment of time the students talk in pairs: the (2i - 1)-th man in the line (for the current moment) talks with the (2i)-th one. 

Let's look at this process in more detail. Let's number the people from 1 to 5. Let's assume that the line initially looks as 23154 (person number 2 stands at the beginning of the line). Then, before the shower opens, 2 talks with 3, 1 talks with 5, 4 doesn't talk with anyone. Then 2 enters the shower. While 2 has a shower, 3 and 1 talk, 5 and 4 talk too. Then, 3 enters the shower. While 3 has a shower, 1 and 5 talk, 4 doesn't talk to anyone. Then 1 enters the shower and while he is there, 5 and 4 talk. Then 5 enters the shower, and then 4 enters the shower.

We know that if students i and j talk, then the i-th student's happiness increases by gij and the j-th student's happiness increases by gji. Your task is to find such initial order of students in the line that the total happiness of all students will be maximum in the end. Please note that some pair of students may have a talk several times. In the example above students 1 and 5 talk while they wait for the shower to open and while 3 has a shower.

Input

The input consists of five lines, each line contains five space-separated integers: the j-th number in the i-th line shows gij (0 ≤ gij ≤ 105). It is guaranteed that gii = 0 for all i.

Assume that the students are numbered from 1 to 5.

Output

Print a single integer — the maximum possible total happiness of the students.

Examples

Input

0 0 0 0 9
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
7 0 0 0 0


Output

32


Input

0 43 21 18 2
3 0 21 11 65
5 2 0 1 4
54 62 12 0 99
87 64 81 33 0


Output

620

Note

In the first sample, the optimal arrangement of the line is 23154. In this case, the total happiness equals:

(g23 + g32 + g15 + g51) + (g13 + g31 + g54 + g45) + (g15 + g51) + (g54 + g45) = 32.
// Main Code at the Bottom
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger; 
public class Main {
	//Fast IO class
    static class FastReader {
        BufferedReader br; 
        StringTokenizer st; 
        public FastReader() {
        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
        	if(!env) {
        		try {
					br=new BufferedReader(new FileReader("src\\input.txt"));
				} catch (FileNotFoundException e) {
					e.printStackTrace();
				}
        	}
        	else br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) {
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
        int nextInt() {
            return Integer.parseInt(next()); 
        } 
        long nextLong() {
            return Long.parseLong(next()); 
        } 
        double nextDouble() {
            return Double.parseDouble(next()); 
        } 
        String nextLine() {
            String str = ""; 
            try {
                str = br.readLine(); 
            } 
            catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }     
    static long MOD=1000000000+7;
    //Euclidean Algorithm
    static long gcd(long A,long B){
    	return (B==0)?A:gcd(B,A%B);
    }
    //Modular Exponentiation
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
    }
    //Modular Inverse
    static long inverse(long x) {
    	return fastExpo(x,MOD-2);
    }
    //Prime Number Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i*i<=n;i+=6) if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
    //Reverse an array
    static void reverse(char arr[],int l,int r){
    	while(l<r) {
    		char tmp=arr[l];
    		arr[l++]=arr[r];
    		arr[r--]=tmp;
    	}
    }
    //Print array
    static void print1d(int arr[]) {
    	out.println(Arrays.toString(arr));
    }
    static void print2d(int arr[][]) {
    	for(int a[]: arr) out.println(Arrays.toString(a));
    }
    // Pair
    static class pair{
    	int x,y;
    	pair(int a,int b){
    		this.x=a;
    		this.y=b;
    	}
    	public boolean equals(Object obj) {
    		if(obj == null || obj.getClass()!= this.getClass()) return false;
            pair p = (pair) obj;
            return (this.x==p.x && this.y==p.y);
        }
    	public int hashCode() {
            return Objects.hash(x,y);
        }
    }
    static FastReader sc=new FastReader();
    static PrintWriter out=new PrintWriter(System.out);  
    //Main function(The main code starts from here
    //Generating next lexicographically greater permutation
    static boolean next_perm(int a[],int n) {
    	if(n<=1) return false;
    	//We only change the position of a "digit" when everything to the right is in descending order
    	//This digit is called pivot
    	int pivot=n-2;
    	while(pivot>=0) {
    		if(a[pivot]<a[pivot+1]) break;
    		pivot--;
    	}
    	//If there is no pivot, it means that the given permutation is already the last permutation
    	if(pivot<0) return false;
    	//Finding rightmost successor to the pivot i.e. the element which is greater than the pivot and farthest to it
    	int next=n-1;
    	for(int i=n-1;i>pivot;i--) {
    		if(a[i]>a[pivot]) {
    			next=i;
    			break;
    		}
    	}
    	//Swapping pivot with next
    	int tmp=a[pivot];
    	a[pivot]=a[next];
    	a[next]=tmp;
    	//Reversing the suffix
    	int l=pivot+1,r=n-1;
    	while(l<r) {
    		tmp=a[l];
    		a[l++]=a[r];
    		a[r--]=tmp;
    	}
    	return true;
    }
    static int a[][],n;
    static int calc(int p[]) {
    	return (a[p[0]][p[1]]+a[p[1]][p[0]]+a[p[2]][p[3]]+a[p[3]][p[2]])+(a[p[1]][p[2]]+a[p[2]][p[1]]+a[p[3]][p[4]]+a[p[4]][p[3]])+(a[p[2]][p[3]]+a[p[3]][p[2]])+(a[p[3]][p[4]]+a[p[4]][p[3]]);
    }
    public static void main (String[] args) throws java.lang.Exception {
    	int test=1;
    	//test=sc.nextInt();
    	while(test-->0){
    		n=5;a=new int[n][n];
    		for(int i=0;i<n;i++) for(int j=0;j<n;j++) a[i][j]=sc.nextInt();
    		int perm[]= {0,1,2,3,4};
    		int max=calc(perm);
    		while(next_perm(perm,n)) max=Math.max(max,calc(perm));
    		out.println(max);
    	}
        out.flush();
        out.close();
    }
}