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: Wrong Answer)

/*
Author : N_o_o_B
Created : December 08 2019 17:56:27 
*/
#include <bits/stdc++.h>
using namespace std;
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
*/
#define TRACE
#ifdef TRACE
#define trace(...) { cerr<<"[ "; __trace__(#__VA_ARGS__, __VA_ARGS__);}
#undef endl
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const pair<Arg1,Arg2> &x) {
    return out<<"("<<x.first<<","<<x.second<<")";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const vector<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const set<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const map<Arg1,Arg2> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
void __trace__(const string name, Arg1&& arg1){
    cerr << name << " : " << arg1 << " ] " << endl;
}
 
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1&& arg1, Args&&... args){
    const string name = names.substr(0,names.find(','));
    cerr<<name<<" : "<<arg1<<" | ";
    __trace__(names.substr(1+(int)name.size()), args...);
}
#else
    #define trace(args...)
#endif
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<pair<int,int>> vii;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<pair<ll,ll>> vll;
typedef vector<vl> vvl;
//typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> oset;
#define fori(i,n) for(int i=0;i<n;i++)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define ford(i,n)  for(int i=n-1;i>=0;i--)
#define trav(x,a) for(auto& x:a)
#define all(x) x.begin(),x.end()
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define fi first
#define se second
clock_t time_p=clock();
void time_taken()
{
    time_p=clock()-time_p;
    cerr<<"Time Taken : "<<(float)(time_p)/CLOCKS_PER_SEC<<"\n";
}
//const ll mod=998244353;
const ll mod=1e9+7;
const ll INF=1e18;
vector<vector<vector<int>>> dp(80,vector<vector<int>>(80,vector<int>(501,-1)));
vvi a;vvi b;
int n,m;
bool f(int i,int j,int sum){
    if(i==n or j==m or sum > 500){
        return 0;
    }
    if(i==n-1 and j==m-1){
        if(sum == a[i][j]-b[i][j] or sum == b[i][j]-a[i][j]) return 1;
        return 0;
    }
    if(dp[i][j][sum]!=-1) return dp[i][j][sum];
    bool ans = 0;
    ans |= f(i+1,j,abs(sum+a[i][j]-b[i][j]));
    ans |= f(i+1,j,abs(sum-a[i][j]+b[i][j]));
    ans |= f(i,j+1,abs(sum+a[i][j]-b[i][j]));
    ans |= f(i,j+1,abs(sum-a[i][j]+b[i][j]));
    dp[i][j][sum]=ans;
    return ans;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    cout.precision(12);cout << fixed;
    cin >> n >> m;
    a=vvi(n,vi(m));
    b=vvi(n,vi(m));
    fori(i,n){
        fori(j,m){
            cin >> a[i][j];
        }
    }
    fori(i,n){
        fori(j,m){
            cin >> b[i][j];
        }
    }
    for(int sum = 0;sum <= (n+m)*80;sum++){
        if(f(0,0,sum)){
            cout << sum << endl;
            return 0;
        }
    }
   
    time_taken();
    return 0;
}