#include <bits/stdc++.h>
#warning That's the baby, that's not my baby
 
typedef long long ll;
 
using namespace std;
 
/**
 
Solutie:
 
Rotim la 45, acum ne fixam unul dintre cele K puncte.
Observam ca va face cel mult n + m - 1 (< 2 * max(n, m)) pasi =>
ne vom uita pentru r pasi, care este intaltimea maxima la care putem ajunge.
Practic avem un query de maxim pe un patrat din matrice => putem face rmq
 
Complexitate temporala: O(N^2 * logN + K * N)
 
**/
 
const int NMAX = 750;
int a[NMAX + 1][NMAX + 1];
int b[2 * NMAX + 2][2 * NMAX + 2];
int rmqLine[12][2 * NMAX + 1][2 * NMAX + 1];
int rmqCol[12][2 * NMAX + 1][2 * NMAX + 1];
int lg2[2 * NMAX + 1];
 
int N;
const int p2[14] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192};
 
int queryLine (int x, int l, int r) {
  l = max(l, 1);
  r = min(r, N);
  int k = lg2[r - l + 1];
  return max(rmqLine[k][x][l], rmqLine[k][x][r - p2[k] + 1]);
}
 
int queryCol (int x, int l, int r) {
  l = max(l, 1);
  r = min(r, N);
  int k = lg2[r - l + 1];
  return max(rmqCol[k][x][l], rmqCol[k][x][r - p2[k] + 1]);
 
}
 
int answer[NMAX * NMAX + 1];
int mini[2 * NMAX + 1];
 
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
 
  int n, m, k;
  cin >> n >> m >> k;
 
  N = max(n, m);
  int maxel = 0;
 
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
      b[i + j - 1][i - j + N] = a[i][j];
      maxel = std::max(maxel, a[i][j]);
    }
  }
 
  N <<= 1;
  for (int i = 2; i <= N; i++) {
    lg2[i] = lg2[i >> 1] + 1;
  }
 
 
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      if (!b[i][j]) {
        rmqLine[0][i][j] = rmqCol[0][j][i] = 1e9;
      }
      rmqLine[0][i][j] = rmqCol[0][j][i] = b[i][j];
//            cout << setw(3) << b[i][j];
    }
//        cout << '\n';
  }
 
  for (int k = 1; p2[k] <= N; k++) {
    for (int i = 1; i <= N; i++) {
      for (int j = 1; j + p2[k] - 1 <= N; j++) {
        rmqLine[k][i][j] = max(rmqLine[k - 1][i][j], rmqLine[k - 1][i][j + p2[k - 1]]);
      }
    }
    for (int j = 1; j <= N; j++) {
      for (int i = 1; i + p2[k] - 1 <= N; i++) {
        rmqCol[k][j][i] = max(rmqCol[k - 1][j][i], rmqCol[k - 1][j][i + p2[k - 1]]);
      }
    }
  }
 
  for (int r = 1; r <= N; r++) {
    mini[r] = 1e9;
  }
 
  while (k--) {
    int x, y;
    cin >> x >> y;
    int i = x + y - 1;
    int j = x - y + N / 2;
//        cout << " ? " << b[i][j] << " - " << a[x][y] << '\n';
    int cur = 0;
    for (int r = 1; r <= N; r++) {
      int x1 = i - r + 1, y1 = j - r + 1, x2 = i + r - 1, y2 = j + r - 1;
      if (x1 > 0) {
        cur = max(cur, queryLine(x1, y1, y2));
      }
      if (x2 <= N) {
        cur = max(cur, queryLine(x2, y1, y2));
      }
      if (y1 > 0) {
        cur = max(cur, queryCol(y1, x1, x2));
      }
      if (y2 <= N) {
        cur = max(cur, queryCol(y2, x1, x2));
      }
      mini[r] = min(mini[r], cur);
    }
  }
 
  for (int r = 1; r <= N; r++) {
//        cout << " > " << mini[r] << '\n';
    if (mini[r] <= n * m) {
      answer[mini[r] + 1] = max(answer[mini[r] + 1], r);
    }
  }
  for (int i = 1; i <= n * m; i++) {
    answer[i] = max(answer[i], answer[i - 1]);
    if (i > maxel) {
      std::cout << "-1 ";
    } else {
      cout << answer[i] << ' ';
    }
  }
 
  return 0;
}
 
 
/**
 
6 5 6
30 14 11 22 16
7  5  6  3  23
20 1  5  2  17
21 1  4  2  6
17 7  3  24 3
26 25 13 14 10
2 2
3 2
4 4
5 5
2 4
4 3
 
**/