#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;
}();

// #define DEBUGx
#ifdef DEBUGx
#define TRC_PG(fmt, args...) fprintf(stderr, "\033[1;32m  TRC_PG(%s:%d):\t\033[0m" fmt, __func__, __LINE__, ##args)
#define TRC_PR(fmt, args...) fprintf(stderr, "\033[1;31m  TRC_PR(%s:%d):\t\033[0m" fmt, __func__, __LINE__, ##args)
#define debug fprintf(stderr, "Passing [%s] in LINE %d\t\n", __func__, __LINE__)
#endif
#ifndef DEBUGx
#define TRC_PG(...)
#define TRC_PR(...)
#define debug
#endif

// 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]));
                TRC_PG("i = %lld, j = %lld, k = %lld, %lld %lld, %lld %lld, %lld %lld, %lld %lld\n", i, j, k, i, j, i - (1 << (k - 1)), j - (1 << (k - 1)), i - (1 << (k - 1)), j, i, j - (1 << (k - 1)));
                TRC_PR("%lld, %lld, %lld, %lld\n", dpMax[i][j][k - 1], dpMax[i - (1 << (k - 1))][j - (1 << (k - 1))][k - 1], dpMax[i - (1 << (k - 1))][j][k - 1], dpMax[i][j - (1 << (k - 1))][k - 1]);
                TRC_PR("%lld %lld\n", dpMin[i][j][k], dpMax[i][j][k]);
            }
        }
    }
    debug;
    long long x = log2(n);
    if ((1 << x) != n)
        x++;
    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]));
            TRC_PG("i = %lld, j = %lld, a = %lld, b = %lld\n", i, j, a, b);
            ans = min(ans, b - a);
        }
    }
}

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

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