/*
方程并不难，主要通过这个题学一下单调队列
*/
#include <cstdio>
#define inf 1000000007
int a[202][33][33],f[2][33][33][202],q[33],qc[33],v[33][202],cnt,n,C,W,T,i,j,k,t,ii,jj,kk,tt,l,r,ans,ch,tag;void read(int &x)
{for (ch=getchar(); ch<=32; ch=getchar());for (x=0; ch>32; ch=getchar()) x = x*10+ch-48;
}void update(int &x, int y)
{if (y > x) x = y;
}void clear()
{l=1; r=0;q[0] = inf;q[1] = -inf;cnt = 0;
}void push(int x)
{q[++r] = x;qc[r] = 1;q[r+1] = -inf;while (q[r] >= q[r-1]){qc[r-1] += qc[r];q[r-1] = q[r];q[r--] = -inf;}if (++cnt > C) if (--qc[l] == 0) q[l++] = inf;
}int main()
{freopen("coin.in", "r", stdin);freopen("coin.out", "w", stdout);scanf("%d%d%d%d", &n, &C, &W, &T);for (t=1; t<=T; ++t)for (i=1; i<=n; ++i)for (j=1; j<=n; ++j) read(a[t][i][j]);for (i=1; i<=n; ++i)for (j=1; j<=n; ++j) f[0][i][j][0] = a[1][i][j];t = 0;for (tt=2; tt<=T; ++tt){t ^= 1;for (i=1; i<=n; ++i)for (j=1; j<=n; ++j)for (k=0; k<=W; ++k) f[t][i][j][k] = -inf;for (i=1; i<=n; ++i){++tag;for (jj=1; jj<=n; ++jj)for (kk=0; kk<=W; ++kk)if (v[jj][kk] != tag){clear();j = jj;k = kk;while (j<=n && k<=W){v[j][k] = tag;update(f[t][i][j][k], q[l]);push(f[t^1][i][j][k]);j += 2;k++;}}++tag;for (jj=n; jj>=1; --jj)for (kk=0; kk<=W; ++kk)if (v[jj][kk] != tag){clear();j = jj;k = kk;while (j>=1 && k<=W){v[j][k] = tag;update(f[t][i][j][k], q[l]);push(f[t^1][i][j][k]);j -= 2;k++;}}}for (j=1; j<=n; ++j){++tag;for (ii=1; ii<=n; ++ii)for (kk=0; kk<=W; ++kk)if (v[ii][kk] != tag){clear();i = ii;k = kk;while (i<=n && k<=W){v[i][k] = tag;update(f[t][i][j][k], q[l]);push(f[t^1][i][j][k]);i += 2;k++;}}++tag;for (ii=n; ii>=1; --ii)for (kk=0; kk<=W; ++kk)if (v[ii][kk] != tag){clear();i = ii;k = kk;while (i>=1 && k<=W){v[i][k] = tag;update(f[t][i][j][k], q[l]);push(f[t^1][i][j][k]);i -= 2;k++;}}}for (i=1; i<=n; ++i)for (j=1; j<=n; ++j)for (k=0; k<=W; ++k){update(f[t][i][j][k], f[t^1][i-1][j][k]);update(f[t][i][j][k], f[t^1][i+1][j][k]);update(f[t][i][j][k], f[t^1][i][j-1][k]);update(f[t][i][j][k], f[t^1][i][j+1][k]);update(f[t][i][j][k], f[t^1][i][j][k]);f[t][i][j][k] += a[tt][i][j];}}for (i=1; i<=n; ++i)for (j=1; j<=n; ++j)for (k=0; k<=W; ++k) update(ans, f[t][i][j][k]);printf("%d\n", ans);return 0;
}