/*
嗯,脑补二维RMQ fail，挂成暴力分。
一维是维护一个区间，二维是维护一个矩阵。
由于这个题是小正方形，三维数组即可。二维RMQ略麻烦...
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>#define x2 x1+n-1-(1<<m)+1
#define y2 y1+n-1-(1<<m)+1using namespace std;
void read(int &x)
{int f=1;x=0;char c=getchar();while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}while(isdigit(c)){x=(x<<3)+(x<<1)+c-'0';c=getchar();}x*=f;
}void out(int x)
{if(!x){putchar('0');return;}if(x<0){x=~x+1;putchar('-');}char c[30]= {0};while(x)c[++c[0]]=x%10+48,x/=10;while(c[0])putchar(c[c[0]--]);
}
const int inf=1e3+29;
int i,j,k,m;
int g[inf][inf][12];
int f[inf][inf][12];
int x,y,n;
int ans=0x7fffffff;int query(int x1,int y1)
{int maxn=0,minn=0x7fffffff;maxn=max(f[x1][y1][m],f[x2][y2][m]);maxn=max(maxn,f[x1][y2][m]);maxn=max(maxn,f[x2][y1][m]);minn=min(g[x1][y1][m],g[x2][y2][m]);minn=min(minn,g[x1][y2][m]);minn=min(minn,g[x2][y1][m]);return maxn-minn;
}int main()
{read(y);read(x);read(n);m=log(n)/log(2);for(i=1; i<=y; i++)for(j=1; j<=x; j++){read(f[j][i][0]);g[j][i][0]=f[j][i][0];}for(k=1; k<=12; k++)for(i=1; i+(1<<k)-1<=x; i++)for(j=1; j+(1<<k)-1<=y; j++){f[i][j][k]=max(f[i][j][k-1],f[i+(1<<(k-1))][j+(1<<(k-1))][k-1]);f[i][j][k]=max(f[i][j][k],f[i+(1<<(k-1))][j][k-1]);f[i][j][k]=max(f[i][j][k],f[i][j+(1<<(k-1))][k-1]);g[i][j][k]=min(g[i][j][k-1],g[i+(1<<(k-1))][j+(1<<(k-1))][k-1]);g[i][j][k]=min(g[i][j][k],g[i+(1<<(k-1))][j][k-1]);g[i][j][k]=min(g[i][j][k],g[i][j+(1<<(k-1))][k-1]);}for(i=1; i+n-1<=x; i++)for(j=1; j+n-1<=y; j++)ans=min(ans,query(i,j));out(ans);return 0;
}