Alice lives on a flat planet that can be modeled as a square grid of size n × n, with rows and columns enumerated from 1 to n. We represent the cell at the intersection of row r and column c with ordered pair (r, c). Each cell in the grid is either land or water.

<image> An example planet with n = 5. It also appears in the first sample test.

Alice resides in land cell (r_1, c_1). She wishes to travel to land cell (r_2, c_2). At any moment, she may move to one of the cells adjacent to where she is—in one of the four directions (i.e., up, down, left, or right).

Unfortunately, Alice cannot swim, and there is no viable transportation means other than by foot (i.e., she can walk only on land). As a result, Alice's trip may be impossible.

To help Alice, you plan to create at most one tunnel between some two land cells. The tunnel will allow Alice to freely travel between the two endpoints. Indeed, creating a tunnel is a lot of effort: the cost of creating a tunnel between cells (r_s, c_s) and (r_t, c_t) is (r_s-r_t)^2 + (c_s-c_t)^2.

For now, your task is to find the minimum possible cost of creating at most one tunnel so that Alice could travel from (r_1, c_1) to (r_2, c_2). If no tunnel needs to be created, the cost is 0.

Input

The first line contains one integer n (1 ≤ n ≤ 50) — the width of the square grid.

The second line contains two space-separated integers r_1 and c_1 (1 ≤ r_1, c_1 ≤ n) — denoting the cell where Alice resides.

The third line contains two space-separated integers r_2 and c_2 (1 ≤ r_2, c_2 ≤ n) — denoting the cell to which Alice wishes to travel.

Each of the following n lines contains a string of n characters. The j-th character of the i-th such line (1 ≤ i, j ≤ n) is 0 if (i, j) is land or 1 if (i, j) is water.

It is guaranteed that (r_1, c_1) and (r_2, c_2) are land.

Output

Print an integer that is the minimum possible cost of creating at most one tunnel so that Alice could travel from (r_1, c_1) to (r_2, c_2).

Examples

Input


5
1 1
5 5
00001
11111
00111
00110
00110


Output


10


Input


3
1 3
3 1
010
101
010


Output


8

Note

In the first sample, a tunnel between cells (1, 4) and (4, 5) should be created. The cost of doing so is (1-4)^2 + (4-5)^2 = 10, which is optimal. This way, Alice could walk from (1, 1) to (1, 4), use the tunnel from (1, 4) to (4, 5), and lastly walk from (4, 5) to (5, 5).

In the second sample, clearly a tunnel between cells (1, 3) and (3, 1) needs to be created. The cost of doing so is (1-3)^2 + (3-1)^2 = 8.
import java.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class cf1130c {
    static BufferedReader __in;
    static PrintWriter __out;
    static StringTokenizer input;

    static int n;

    public static void main(String[] args) throws IOException {
        __in = new BufferedReader(new InputStreamReader(System.in));
        __out = new PrintWriter(new OutputStreamWriter(System.out));

        n = ri();
        int begin = rni() * n + ni() - n - 1, end = rni() * n + ni() - n - 1;
        boolean[][] land = new boolean[n][n];
        DSU dsu = new DSU(n * n);
        for(int i = 0; i < n; ++i) {
            char[] row = rcha();
            for(int j = 0; j < n; ++j) {
                if(land[i][j] = row[j] == '0') {
                    int ind = i * n + j;
                    dsu.make(ind);
                    if(i > 0 && dsu.par[(i - 1) * n + j] >= 0) {
                        dsu.union((i - 1) * n + j, ind);
                    }
                    if(j > 0 && dsu.par[i * n + j - 1] >= 0) {
                        dsu.union(i * n + j - 1, ind);
                    }
                }
            }
        }
        if(dsu.find(begin) == dsu.find(end)) {
            prln(0);
        } else {
            int ans = IMAX;
            for(P a : dsu.set.get(dsu.find(begin))) {
                // prln(a);
                for(P b : dsu.set.get(dsu.find(end))) {
                    // pr("\t");
                    // prln(b);
                    int dx = b.a - a.a, dy = b.b - a.b;
                    ans = min(ans, dx * dx + dy * dy);
                }
            }
            prln(ans);
        }

        close();
    }

    static class P {
        int a, b;

        P(int a_, int b_) {
            a = a_;
            b = b_;
        }

        @Override
        public String toString() {
            return "Pair{" + "a = " + a + ", b = " + b + '}';
        }

        public boolean equalsSafe(Object o) {
            if(this == o) return true;
            if(o == null || getClass() != o.getClass()) return false;
            P p = (P) o;
            return a == p.a && b == p.b;
        }

        public boolean equalsUnchecked(Object o) {
            P p = (P) o;
            return a == p.a && b == p.b;
        }

        @Override
        public boolean equals(Object o) {
            return equalsUnchecked(o);
        }

        @Override
        public int hashCode() {
            return Objects.hash(a, b);
        }
    }

    static class DSU {
        int[] par;
        List<Set<P>> set = new ArrayList<>();

        DSU(int n) {
            par = new int[n];
            for(int i = 0; i < n; ++i) {
                set.add(new HashSet<>());
            }
            fill(par, -1);
        }

        void make(int v) {
            par[v] = v;
            set.get(v).add(new P(v / n, v % n));
        }

        int find(int v) {
            if(par[v] == -1) {
                return -1;
            } else if(v == par[v]) {
                return v;
            } else {
                par[v] = find(par[v]);
                return par[v];
            }
        }

        void union(int u, int v) {
            int a = find(u), b = find(v);
            if(a != b) {
                if(b < a) {
                    int swap = a;
                    a = b;
                    b = swap;
                }
                par[b] = a;
                set.get(a).addAll(set.get(b));
            }
        }
    }

    // references
    // IBIG = 1e9 + 7
    // IRAND ~= 3e8
    // IMAX ~= 2e10
    // LMAX ~= 9e18
    // constants
    static final int IBIG = 1000000007;
    static final int IRAND = 327859546;
    static final int IMAX = 2147483647;
    static final int IMIN = -2147483648;
    static final long LMAX = 9223372036854775807L;
    static final long LMIN = -9223372036854775808L;

    // util
    static int minof(int a, int b, int c) {
        return min(a, min(b, c));
    }

    static int minof(int... x) {
        return x.length == 1 ? x[0] : x.length == 2 ? min(x[0], x[1]) : min(x[0], minstarting(1, x));
    }

    static int minstarting(int offset, int... x) {
        assert x.length > 2;
        return offset == x.length - 2 ? min(x[offset], x[offset + 1]) : min(x[offset], minstarting(offset + 1, x));
    }

    static long minof(long a, long b, long c) {
        return min(a, min(b, c));
    }

    static long minof(long... x) {
        return x.length == 1 ? x[0] : x.length == 2 ? min(x[0], x[1]) : min(x[0], minstarting(1, x));
    }

    static long minstarting(int offset, long... x) {
        assert x.length > 2;
        return offset == x.length - 2 ? min(x[offset], x[offset + 1]) : min(x[offset], minstarting(offset + 1, x));
    }

    static int maxof(int a, int b, int c) {
        return max(a, max(b, c));
    }

    static int maxof(int... x) {
        return x.length == 1 ? x[0] : x.length == 2 ? max(x[0], x[1]) : max(x[0], maxstarting(1, x));
    }

    static int maxstarting(int offset, int... x) {
        assert x.length > 2;
        return offset == x.length - 2 ? max(x[offset], x[offset + 1]) : max(x[offset], maxstarting(offset + 1, x));
    }

    static long maxof(long a, long b, long c) {
        return max(a, max(b, c));
    }

    static long maxof(long... x) {
        return x.length == 1 ? x[0] : x.length == 2 ? max(x[0], x[1]) : max(x[0], maxstarting(1, x));
    }

    static long maxstarting(int offset, long... x) {
        assert x.length > 2;
        return offset == x.length - 2 ? max(x[offset], x[offset + 1]) : max(x[offset], maxstarting(offset + 1, x));
    }

    static int powi(int a, int b) {
        if(a == 0) return 0;
        int ans = 1;
        while(b > 0) {
            if((b & 1) > 0) ans *= a;
            a *= a;
            b >>= 1;
        }
        return ans;
    }

    static long powl(long a, int b) {
        if(a == 0) return 0;
        long ans = 1;
        while(b > 0) {
            if((b & 1) > 0) ans *= a;
            a *= a;
            b >>= 1;
        }
        return ans;
    }

    static int floori(double d) {
        return (int) d;
    }

    static int ceili(double d) {
        return (int) ceil(d);
    }

    static long floorl(double d) {
        return (long) d;
    }

    static long ceill(double d) {
        return (long) ceil(d);
    }

    // input
    static void r() throws IOException {
        input = new StringTokenizer(__in.readLine());
    }

    static int ri() throws IOException {
        return Integer.parseInt(__in.readLine());
    }

    static long rl() throws IOException {
        return Long.parseLong(__in.readLine());
    }

    static int[] ria(int n) throws IOException {
        int[] a = new int[n];
        input = new StringTokenizer(__in.readLine());
        for(int i = 0; i < n; ++i) a[i] = Integer.parseInt(input.nextToken());
        return a;
    }

    static long[] rla(int n) throws IOException {
        long[] a = new long[n];
        input = new StringTokenizer(__in.readLine());
        for(int i = 0; i < n; ++i) a[i] = Long.parseLong(input.nextToken());
        return a;
    }

    static char[] rcha() throws IOException {
        return __in.readLine().toCharArray();
    }

    static String rline() throws IOException {
        return __in.readLine();
    }

    static int rni() throws IOException {
        input = new StringTokenizer(__in.readLine());
        return Integer.parseInt(input.nextToken());
    }

    static int ni() {
        return Integer.parseInt(input.nextToken());
    }

    static long rnl() throws IOException {
        input = new StringTokenizer(__in.readLine());
        return Long.parseLong(input.nextToken());
    }

    static long nl() {
        return Long.parseLong(input.nextToken());
    }

    // output
    static void pr(int i) {
        __out.print(i);
    }

    static void prln(int i) {
        __out.println(i);
    }

    static void pr(long l) {
        __out.print(l);
    }

    static void prln(long l) {
        __out.println(l);
    }

    static void pr(double d) {
        __out.print(d);
    }

    static void prln(double d) {
        __out.println(d);
    }

    static void pr(char c) {
        __out.print(c);
    }

    static void prln(char c) {
        __out.println(c);
    }

    static void pr(char[] s) {
        __out.print(new String(s));
    }

    static void prln(char[] s) {
        __out.println(new String(s));
    }

    static void pr(String s) {
        __out.print(s);
    }

    static void prln(String s) {
        __out.println(s);
    }

    static void pr(Object o) {
        __out.print(o);
    }

    static void prln(Object o) {
        __out.println(o);
    }

    static void prln() {
        __out.println();
    }

    static void pryes() {
        __out.println("yes");
    }

    static void pry() {
        __out.println("Yes");
    }

    static void prY() {
        __out.println("YES");
    }

    static void prno() {
        __out.println("no");
    }

    static void prn() {
        __out.println("No");
    }

    static void prN() {
        __out.println("NO");
    }

    static void pryesno(boolean b) {
        __out.println(b ? "yes" : "no");
    }

    ;

    static void pryn(boolean b) {
        __out.println(b ? "Yes" : "No");
    }

    static void prYN(boolean b) {
        __out.println(b ? "YES" : "NO");
    }

    static void prln(int... a) {
        for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i) ;
        __out.println(a[a.length - 1]);
    }

    static void prln(long... a) {
        for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i) ;
        __out.println(a[a.length - 1]);
    }

    static <T> void prln(Collection<T> c) {
        int n = c.size() - 1;
        Iterator<T> iter = c.iterator();
        for(int i = 0; i < n; __out.print(iter.next()), __out.print(' '), ++i) ;
        __out.println(iter.next());
    }

    static void flush() {
        __out.flush();
    }

    static void close() {
        __out.close();
    }
}