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

static auto __fast_io = []()
{
    ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    return 0;
}();

// 有一个 a*b 的整数组成的矩阵，现请你从中找出一个 n*n 的正方形区域，使得该区域所有数中的最大值和最小值的差最小。

const long long MAX = 1000 + 10;
const long long INF = 1e9 + 10;

#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))

long long dpMin[MAX][MAX][30]; // 表示 [i - (1 << k)][j - (i << k)] 到 [i][j] 的最小值
long long dpMax[MAX][MAX][30];
long long ans;
long long p[MAX][MAX];
long long a, b, n;

void input()
{
    cin >> a >> b >> n;
    for (long long i = 1; i <= a; i++)
    {
        for (long long j = 1; j <= b; j++)
        {
            cin >> p[i][j];
        }
    }
}

void work()
{
    for (long long i = 1; i <= a; i++)
    {
        dpMin[i][0][0] = INF;
    }
    for (long long j = 1; j <= b; j++)
    {
        dpMin[0][j][0] = INF;
    }
    for (long long i = 1; i <= a; i++)
    {
        for (long long j = 1; j <= b; j++)
        {
            dpMin[i][j][0] = p[i][j];
            dpMax[i][j][0] = p[i][j];

            for (long long k = 1; (1 << k) <= i + 1 && (1 << k) <= j + 1; k++)
            {
                dpMin[i][j][k] = min(min(dpMin[i][j][k - 1], dpMin[i - (1 << (k - 1))][j - (1 << (k - 1))][k - 1]),
                                     min(dpMin[i - (1 << (k - 1))][j][k - 1], dpMin[i][j - (1 << (k - 1))][k - 1]));
                dpMax[i][j][k] = max(max(dpMax[i][j][k - 1], dpMax[i - (1 << (k - 1))][j - (1 << (k - 1))][k - 1]),
                                     max(dpMax[i - (1 << (k - 1))][j][k - 1], dpMax[i][j - (1 << (k - 1))][k - 1]));
            }
        }
    }
    long long x = log2(n);
    ans = INF;
    for (long long i = n; i <= a; i++)
    {
        for (long long j = n; j <= b; j++)
        {
            long long a = min(min(dpMin[i][j][x - 1], dpMin[i - (1 << (x - 1))][j - (1 << (x - 1))][x - 1]),
                              min(dpMin[i - (1 << (x - 1))][j][x - 1], dpMin[i][j - (1 << (x - 1))][x - 1]));
            long long b = max(max(dpMax[i][j][x - 1], dpMax[i - (1 << (x - 1))][j - (1 << (x - 1))][x - 1]),
                              max(dpMax[i - (1 << (x - 1))][j][x - 1], dpMax[i][j - (1 << (x - 1))][x - 1]));
            // cout << i << " " << j << " " << a << " " << b << "\n";
            ans = min(ans, b - a);
        }
    }
}

void output()
{
    cout << ans << "\n";
}

int main()
{
    input();
    work();
    output();
    return 0;
}