Problem:
Score : 500 points

Problem Statement

We have a grid with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left.

The square (i, j) has two numbers A_{ij} and B_{ij} written on it.

First, for each square, Takahashi paints one of the written numbers red and the other blue.

Then, he travels from the square (1, 1) to the square (H, W). In one move, he can move from a square (i, j) to the square (i+1, j) or the square (i, j+1). He must not leave the grid.

Let the unbalancedness be the absolute difference of the sum of red numbers and the sum of blue numbers written on the squares along Takahashi's path, including the squares (1, 1) and (H, W).

Takahashi wants to make the unbalancedness as small as possible by appropriately painting the grid and traveling on it.

Find the minimum unbalancedness possible.

Constraints

2 \leq H \leq 80

2 \leq W \leq 80

0 \leq A_{ij} \leq 80

0 \leq B_{ij} \leq 80

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H W
A_{11} A_{12} \ldots A_{1W}
:
A_{H1} A_{H2} \ldots A_{HW}
B_{11} B_{12} \ldots B_{1W}
:
B_{H1} B_{H2} \ldots B_{HW}

Output

Print the minimum unbalancedness possible.

Sample Input 1

2 2
1 2
3 4
3 4
2 1

Sample Output 1

0

By painting the grid and traveling on it as shown in the figure below, the sum of red numbers and the sum of blue numbers are 3+3+1=7 and 1+2+4=7, respectively, for the unbalancedness of 0.

Sample Input 2

2 3
1 10 80
80 10 1
1 2 3
4 5 6

Sample Output 2

2

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

#include<bits/stdc++.h>
using namespace std;
int dp[81][81][6400 * 2 + 1];
// because max distance that we can travel is to first go form start, go to last column in first row and then in last column, go till last
// row till we reach our destination, all other path distance will be less as they kind of form diagnal path.
int find_it(int dif_sum , int i , int j , int &n , int &m , vector<vector<pair<int , int>>> & matrix){
    if(i == n - 1 && j == m - 1){
        return min(abs(dif_sum + matrix[i][j].first - matrix[i][j].second) , abs(dif_sum + matrix[i][j].second - matrix[i][j].first));
    }
    if(dp[i][j][dif_sum] != -1){
        return dp[i][j][dif_sum];
    }
    int minvalue = INT_MAX;
    int ff = abs(dif_sum + matrix[i][j].first - matrix[i][j].second);
    int ss = abs(dif_sum + matrix[i][j].second - matrix[i][j].first);
    // abs because suppose till i & j , we have got k sum and -k sum possible, what is the meaning of keeping of both k and -k as from now on 
    // they both are going to have same all possible path till n - 1 & m - 1, so so reduce time , we can just use one value that is k
    // and as problem mentions we only care about dif , so it means absolute
    if(i + 1 < n){
        minvalue = min(minvalue , find_it(ff , i + 1 , j , n , m , matrix));
        minvalue = min(minvalue , find_it(ss , i + 1 , j , n , m , matrix));

    }
    if(j + 1 < m){
        minvalue = min(minvalue , find_it(ff , i , j + 1 , n , m , matrix));
        minvalue = min(minvalue , find_it(ss , i , j + 1 , n , m , matrix));
    }
    return dp[i][j][dif_sum] = minvalue;

}
int main(){

	ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n , m;
    cin >> n >> m;
    vector<vector<pair<int , int> > > matrix(n , vector<pair<int , int>> (m));

    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m; ++j){
            cin >> matrix[i][j].first;
        }
    }
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m; ++j){
            cin >> matrix[i][j].second;
        }
    }
    memset(dp , -1 , sizeof(dp));
    cout << find_it(0 , 0 , 0 , n , m , matrix);


return 0;
}