#include <bits/stdc++.h>
#define ll long long
#define N 101
#define inf -10001
using namespace std;

vector<int> edges[N * N];
int R, C;
int a[N][N];
int dp[N];

int main() {
	scanf("%d%d", &R, &C);
	for (int i = 1; i <= R; i++)
		for (int j = 1; j <= C; j++)
			scanf("%d", &a[i][j]);
	for (int i = 1; i <= R; i++)
		for (int j = 1; j <= C; j++) {
			if (a[i][j] > a[i + 1][j] && i + 1 <= R)//有向边是从编号大的点向编号小的点连
				edges[a[i][j]].push_back(a[i + 1][j]);
			if (a[i][j] > a[i - 1][j] && i - 1 >= 1)
				edges[a[i][j]].push_back(a[i - 1][j]);
			if (a[i][j] > a[i][j + 1] && j + 1 <= C)
				edges[a[i][j]].push_back(a[i][j + 1]);
			if (a[i][j] > a[i][j - 1] && j - 1 >= 1)
				edges[a[i][j]].push_back(a[i][j - 1]);
		}//建一个有向图
	for (int i = 1; i <= R * C; i++)
		dp[i] = inf;
	dp[1] = 1;
	for (int i = 2; i <= R * C; i++) {
		for (auto &j : edges[i])
			dp[i] = max(dp[i], dp[j] + 1);
	}//转移状态
	int ans = 0;
	for (int i = 1; i <= R * C; i++)
		ans = max(dp[i], ans);//求答案
	printf("%d", ans);
}