// 想定解法: 格子点平面上の最短経路探索(Dijkstra, SPFA, etc.)
//           各位置での累計移動距離ごとの必要コスト最小化
//           もっといい解がありそうな気がするのでつよい人に期待

#include <cstdio>
#include <iostream>
#include <deque>
#include <map>
using namespace std;
#define REP(i, n)           for(int(i)=0;(i)<(n);++(i))

const int MAXN = 100;

int board[MAXN][MAXN];
int cost[MAXN][MAXN];
const int INF = 1<<29;
const int dir[][2] = {{1,0},{0,1},{-1,0},{0,-1}};

int N, V, SX, SY, GX, GY;

int solve(){
    REP(y,N) REP(x,N) cost[y][x] = INF;

    deque<pair<pair<int,int>,int> > q;
    q.push_back(make_pair(make_pair(SX,SY),0));
    cost[SY][SX] = 0;

    while(!q.empty()){
        auto &v = q.front();
        int x = v.first.first, y = v.first.second, s = v.second;
        q.pop_front();
        int nowcost = cost[y][x];
        if(x == GX && y == GY) return s;

        for(int d = 0; d < 4; d++){
            int mx = x + dir[d][0];
            int my = y + dir[d][1];
            if(mx < 0 || my < 0 || mx >= N || my >= N) continue;
            int nextcost = nowcost + board[my][mx];
            if(cost[my][mx] > nextcost && nextcost < V){
                cost[my][mx]= nextcost;
                q.push_back(make_pair(make_pair(mx,my),s+1));
            }
        }
    }
    return -1;
}

int main(){
    cin >> N >> V >> SX >> SY >> GX >> GY;
    REP(y,N) REP(x,N) cin >> board[y][x];
    SX--,SY--,GX--,GY--;

    cout << solve() << endl;
    return 0;
}