A

Problem description

You've got a 5 × 5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:

1. Swap two neighboring matrix rows, that is, rows with indexes i and i + 1 for some integer i (1 ≤ i < 5).
2. Swap two neighboring matrix columns, that is, columns with indexes j and j + 1for some integer j (1 ≤ j < 5).

You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.

Input

The input consists of five lines, each line contains five integers: the j-th integer in the i-th line of the input represents the element of the matrix that is located on the intersection of the i-th row and the j-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.

Output

Print a single integer — the minimum number of moves needed to make the matrix beautiful.

Examples

Input

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

Output

3

Input

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

Output

1
解题思路：题目的意思就是有一个5*5的矩阵，其中只有一个元素值为1，其余元素的值全部为0，从元素1这个位置移到中心点（2，2）（下标从0~4）的过程中，每次只能向上、下、左、右（其中一个方向）移动一步，求最小的移动步数。简单模拟一下过程即可推出：设1元素的坐标为(row,col)，则移动的最小步数为abs(row-2)+abs(col-2)，简单AC。
AC代码：

#include<bits/stdc++.h>
using namespace std;
int main(){
    int row,col,x;
    for(int i=0;i<5;++i){
        for(int j=0;j<5;++j){
            cin>>x;
            if(x){row=i;col=j;}
        }
    }
    cout<<abs(row-2)+abs(col-2)<<endl;
    return 0;
}