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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
#define ll long long
#define ld long double
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repo(i,n) for(int i = 1; i < (int)(n); i++)
#define pb push_back
#define mp make_pair
#define np next_permutation
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define uniq(v) v.erase(unique(v.begin(),v.end()),v.end())
#define lb(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define ub(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
using Pair = pair<ll,pair<int,int>>;
#define pq priority_queue<Pair, vector<Pair>, greater<Pair>> 
const ll mod=1000000007;
//const ll mod=998244353;
const ld pi=acos(-1.0);
const ll INF = 1LL<<61;
template<class T>bool chmax(T &a, const T &b) { 
  if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) {
  if (b<a) { a=b; return 1; } return 0; }
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
//intの最大値2147483647 ≒ 2×10^9
//long longの最大値9223372036854775807 ≒ 9×10^18
//'大文字'+=32;    で小文字に
//  cout << fixed << setprecision (20);   小数点以下2０桁まで
//実行時間制約2秒では２×10^8回くらいまで計算できる




int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int h,w;
  cin>>h>>w;

  vector<vector<int>> p(h,vector<int>(w));
  rep(i,h){
    rep(j,w){
      cin>>p[i][j];
    }
  }

  rep(i,h){
    rep(j,w){
      int x;
      cin>>x;
      p[i][j]=abs(p[i][j]-x);
    }
  }

  bool ok[h][w][161];
  memset(ok,false,sizeof(ok));

  ok[0][0][80+p[0][0]]=true;
  ok[0][0][80-p[0][0]]=true;

  rep(i,h){
    rep(j,w){
      if(i+1<h) {
        int t=p[i+1][j];
        rep(k,161){
          if(ok[i][j][k]) {
            if(k+t<161) ok[i+1][j][k+t]=true;
            if(k-t>=0) ok[i+1][j][k-t]=true;
          }
        }
      }
      if(j+1<w) {
        int t=p[i][j+1];
        rep(k,161){
          if(ok[i][j][k]) {
            if(k+t<161) ok[i][j+1][k+t]=true;
            if(k-t>=0) ok[i][j+1][k-t]=true;
          }
        }
      }
    }
  }

  rep(i,81){
    if(ok[h-1][w-1][80+i]||ok[h-1][w-1][80-i]) {
      cout << i << endl;
      return 0;
    }
  }

}