#include<bits/stdc++.h>
#include<atcoder/all>
#define chmin(x,y) (x) = min((x),(y))
#define chmax(x,y) (x) = max((x),(y))
using namespace std;
using namespace atcoder;
using ll = long long;
const ll mod = 998244353;
using mint = modint998244353;
using Graph = vector<vector<int>>;

const vector<int> dx = {1,0,-1,0}, dy = {0,1,0,-1};

int main(){
  // input
  int N,M,K; cin >> N >> M >> K;
  vector<vector<int>> A(N,vector<int>(M));
  for(int i = 0; i < N; i++)
    for(int j = 0; j < M; j++)
      cin >> A[i][j];
      
  // solve
  int lo = 1, hi = 1e9+1;
  while(hi - lo > 1){
    int mid = (lo + hi) / 2;
    
    deque<tuple<int,int,int>> d;
    d.emplace_back(0,0,int(mid > A[0][0]));
    
    vector<vector<int>> seen(N,vector<int>(M,1e9));
    
    while(true){
      auto[x,y,now] = d.front();
      d.pop_front();
      
      if(x+1 == N && y+1 == M) break;
      
      for(int i = 0; i < 4; i++){
        int nx = x + dx[i], ny = y + dy[i];
        if(nx < 0 || ny < 0 || nx == N || ny == M) continue;
        
        if(A[nx][ny] >= mid && seen[nx][ny] > now){
          d.emplace_front(nx,ny,now);
          seen[nx][ny] = now;
        }
        else if(A[nx][ny] < mid && seen[nx][ny] > now + 1){
          d.emplace_back(nx,ny,now+1);
          seen[nx][ny] = now + 1;
        }
      }
    }
    
    if(seen[N-1][M-1] <= K) lo = mid; else hi = mid;
  }
  
  
  // output
  cout << lo << endl;
}