#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][j] 到 [i + (1 << k)][j + (i << k)] 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++)
    {
        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; k < 30; k++)
    {
        for (long long i = 1; i + (1 << k) - 1 <= a; i++)
        {
            for (long long j = 1; j + (1 << k) - 1 <= b; j++)
            {
                dpMin[i][j][k] = min(min(dpMin[i][j][k - 1], dpMin[i + (1 << (k - 1))][j][k - 1]), min(dpMin[i][j + (1 << (k - 1))][k - 1], dpMin[i + (1 << (k - 1))][j + (1 << (k - 1))][k - 1]));
                dpMax[i][j][k] = max(max(dpMax[i][j][k - 1], dpMax[i + (1 << (k - 1))][j][k - 1]), max(dpMax[i][j + (1 << (k - 1))][k - 1], dpMax[i + (1 << (k - 1))][j + (1 << (k - 1))][k - 1]));
            }
        }
    }

    int x = log2(n);
    ans = INF;
    for (int i = 1; i + n - 1 <= a; i++)
    {
        for (int j = 1; j + n - 1 <= b; j++)
        {
            int a = min(min(dpMin[i][j][x], dpMin[i + n - (1 << x)][j][x]), min(dpMin[i][j + n - (1 << x)][x], dpMin[i + n - (1 << x)][j + n - (1 << x)][x]));
            int b = max(max(dpMax[i][j][x], dpMax[i + n - (1 << x)][j][x]), max(dpMax[i][j + n - (1 << x)][x], dpMax[i + n - (1 << x)][j + n - (1 << x)][x]));
            TRC_PG("i = %d, j = %d, a = %d, b = %d\n", i, j, a, b);
            ans = min(ans, b - a);
        }
    }
}

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

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