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)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define SPEED ios_base::sync_with_stdio(false); cin.tie(NULL);
#define FOR(i, a, b) for (ll i = a; i < b; ++i)
#define RFOR(i, b, a) for (ll i = b; i >= a; --i)
#define ALL(x) x.begin(), x.end()
#define DEBUG(args...) { string _s = #args; replace(ALL(_s), ' ', '\0');\
replace(ALL(_s), ',', ' ');\
stringstream _ss(_s); istream_iterator<string> _it(_ss); _debug(_it, args);}
#define endl "\n"
#define F first
#define S second
#define pb(x) push_back(x)
#define mp(x, y) make_pair(x, y)

using namespace std;
using namespace __gnu_pbds;

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

void _debug(istream_iterator<string>) {}
template<typename T, typename... Args>
void _debug(istream_iterator<string> it, T first, Args... args) {
    cerr << ">> " << *it << " : " << first << endl; _debug(++it, args...);
}
template <typename T1, typename T2>
inline ostream& operator << (ostream& out, const pair<T1, T2>& p) {
    return out << "(" << p.F << ", " << p.S << ")";
}
template<typename T>
inline ostream& operator << (ostream& out, const vector<T>& v) {
    if (v.empty()) return out << "[]";
    else { out << '['; for (auto& e : v) { out << e << ", "; } return out << "\b\b]"; }
}
template<typename T>
inline ostream& operator << (ostream& out, const set<T>& s) {
    if (s.empty()) return out << "{}";
    else { out << '{'; for (auto& e : s) { out << e << ", "; } return out << "\b\b}"; }
}
template<typename T>
inline ostream& operator << (ostream& out, const unordered_set<T>& s) {
    return out << set<T>(ALL(s));
}
template<typename T1, typename T2>
inline ostream& operator << (ostream& out, const map<T1, T2>& m) {
    if (m.empty()) return out << "{}";
    out << '{'; for (auto& p : m) { out << p << ", "; } return out << "\b\b}";
}
template<typename T1, typename T2>
inline ostream& operator << (ostream& out, const unordered_map<T1, T2>& m) {
    return out << map<T1, T2>(ALL(m));
}
template<typename T>
inline ostream& operator << (ostream& out, const ordered_set<T>& s) {
    return out << set<T>(ALL(s));
}

typedef long long ll;
typedef long double ld;
typedef vector<long long> vll;
typedef pair<ll, ll> pll;
typedef vector<pair<ll, ll>> vpll;
typedef unordered_map<ll, ll> STll;
/************************************** MAIN PROGRAM ********************************************/
const int MAX = 81;
int grid[MAX][MAX];
short cache[MAX][MAX][MAX * MAX];
int h, w;

bool ans(int r, int c, int diff) {
    // DEBUG(r, c, diff)
    if ((r == h && c == w - 1) || (r == h - 1 && c == w)) {
        //    DEBUG(diff)
        return diff == 0;
    }
    if (!(r < h && c < w && r >= 0 && c >= 0)) return false;
    if (cache[r][c][diff] != -1) return cache[r][c][diff];


    bool down = ans(r + 1, c, diff - grid[r][c]) | ans(r + 1, c, diff + grid[r][c]);
    bool right = ans(r, c + 1, diff - grid[r][c]) | ans(r, c + 1, diff + grid[r][c]);
    return cache[r][c][diff] = down | right;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    SPEED

    memset(cache, -1, sizeof cache);
    cin >> h >> w;
    FOR(i, 0, h) {
        FOR(j, 0, w) {
            cin >> grid[i][j];
        }
    }
    FOR(i, 0, h) {
        FOR(j, 0, w) {
            int x;
            cin >> x;
            grid[i][j] -= x;
            grid[i][j] = abs(grid[i][j]);
         //   cout << grid[i][j] << ' ';
        }
       // cout << endl;
    }
   // DEBUG(ans(1, 2, 5))
    FOR(res, 0, 2 * MAX) {
        if (ans(0, 0, res)) {
            cout << res;
            return 0;
        }
    }
}
/************************************** END OF PROGRAM ******************************************/
/** Stuff you should look for:
 * int overflow, array bounds, over-counting, graph is not a tree/connected
 * special cases (n=1?), set/unordered_set TLE, multi-set/set error
 * do something instead of nothing and stay organized
 */