#include <bits/stdc++.h>
using namespace std;

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second
#define FILL(a, b) memset(a, b, sizeof(a))

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

typedef long double db;

const int N = 47;
const int N2 = N * N;
const int INF = 1e9;


int a[N2];
int n, m;

int idx(int i, int j){
	return i * m + j;
}


bool ok(int x, int y){
	return 0 <= x && x < n && 0 <= y && y < m;
}

int di[4] = {0, 0, -1, 1};
int dj[4] = {-1, 1, 0, 0};

VI g[N2];
int d[N2][N2];
bool used[N2];


void dfs(int t, int v, int p){
	used[v] = 1;
	for(int to : g[v]){
		if(to == p)
			continue;
		assert(!used[to]);
		
		d[t][to] = d[t][v] + 1;
		dfs(t, to, v);
	}
}


void check(int x, int y){
	int st = idx(x, y);
	FOR(i, 0, n * m){
		g[i].clear();
		d[st][i] = INF;
		used[i] = 0;
	}
	
	FOR(i, 0, n){
		FOR(j, 0, m){
			if(i == x && y == j)
				continue;
			int v = idx(i, j);
			
			int go = -1;
			int dis = INF;
			
			FOR(t, 0, 4){
				int ni = i + di[t];
				int nj = j + dj[t];
				if(!ok(ni, nj))
					continue;
					
				int to = idx(ni, nj);
				int cur = abs(a[st] - a[to]);
				
				if(cur == dis){
					if(abs(a[to] - a[v]) < abs(a[go] - a[v]))
						go = to;
				}
				
				if(cur  < dis){
					dis = cur;
					go = to;
				}
			}
			if(go == -1)
				continue;
			g[go].PB(v);
		}
	}
	d[st][st] = 0;
	dfs(st, st, -1);
	
}





int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	FOR(i, 0, n)
		FOR(j, 0, m)
			cin >> a[idx(i, j)];
	
	//check(1, 1);
	//return 0;
	
	FOR(i, 0, n)
		FOR(j, 0, m)
			check(i, j);
	
	int ans = -1;
	int res = INF;
	
	FOR(kiosk, 0, n * m){
		int mx = 0;
		FOR(other, 0, n * m)
			mx = max(mx, d[other][kiosk]);
		if(mx < res){
			res = mx;
			ans = kiosk;
		}
	}
	if(ans == -1){
		cout << "impossible" << "\n";
		return 0;
	}
	cout << a[ans] << " " << res << endl;
	
	
	return 0;
}