#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

bool isKadomatsu(int a, int b, int c) {
	return a != c && ((a > b && b < c) || (a < b && b > c));
}

int main() {
	int W, H;
	scanf("%d%d", &W, &H);
	vector<vi> M(H, vi(W));
	rep(i, H) rep(j, W) scanf("%d", &M[i][j]);
	static bool vis[100][100][11];
	mset(vis, 0);
	vector<pair<pii,int> > q, nq;
	nq.push_back(mp(mp(0, 0), 10));
	int t = 0;
	while(!nq.empty()) {
		nq.swap(q);
		while(!q.empty()) {
			pii p = q.back().first; int r = q.back().second; q.pop_back();
			if(vis[p.first][p.second][r]) continue;
			vis[p.first][p.second][r] = true;
//			cerr << p.first << ", " << p.second << ", " << r << ": " << t << endl;
			if(p == mp(H-1, W-1)) {
				printf("%d\n", t);
				return 0;
			}
			static const int dy[4] = {0,1,0,-1}, dx[4] = {1,0,-1,0};
			rep(d, 4) {
				int yy = p.first + dy[d], xx = p.second + dx[d];
				if(yy<0||yy>=H||xx<0||xx>=W) continue;
				if(t >= 1 && !isKadomatsu(r, M[p.first][p.second], M[yy][xx])) continue;
				nq.push_back(mp(mp(yy, xx), M[p.first][p.second]));
			}
		}
		++ t;
	}
	puts("-1");
	return 0;
}