#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long infLL = 0x3f3f3f3f3f3f3f3fLL;
const int hash_mod = 1000037;
const double eps = 1e-10;
const double pi = acos(-1);
const int maxn = 2000 + 5;
int n, m;
pair<int, int> a[maxn], b[maxn];
int fa[maxn], fb[maxn];
bitset<maxn> mask[maxn];
int find(int x, int f[]) { return f[x] == -1 ? x : f[x] = find(f[x], f); }
void merge(int x, int y, int f[]) {
  x = find(x, f);
  y = find(y, f);
  if (x != y) f[x] = y;
}
bool check(int t) {
  t *= 2;
  memset(fa, -1, sizeof(fa));
  memset(fb, -1, sizeof(fb));
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j)
      if (a[i].first + a[i].second == a[j].first + a[j].second)
        if (abs(a[i].first - a[j].first) <= t * 2) merge(i, j, fa);
  for (int i = 0; i < m; ++i)
    for (int j = i + 1; j < m; ++j)
      if (b[i].first - b[i].second == b[j].first - b[j].second)
        if (abs(b[i].first - b[j].first) <= t * 2) merge(i, j, fb);
  for (int i = 0; i < n; ++i) mask[i].reset();
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      int sum = a[i].first + a[i].second;
      int sub = b[j].first - b[j].second;
      int x = (sum + sub) / 2, y = (sum - sub) / 2;
      if (abs(a[i].first - x) <= t && abs(b[j].first - x) <= t)
        mask[find(i, fa)].set(find(j, fb));
    }
  for (int i = 0; i < n; ++i) {
    int s = find(i, fa);
    for (int j = i + 1; j < n; ++j) {
      int t = find(j, fa);
      if (t != s && (mask[s] & mask[t]).count() > 1) return true;
    }
  }
  return false;
}
void solve() {
  int l = 1, r = 10000000;
  if (!check(r)) {
    puts("Poor Sereja!");
    return;
  }
  while (l < r) {
    int mid = l + (r - l) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  printf("%d\n", r);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &a[i].first, &a[i].second);
    a[i].first *= 2;
    a[i].second *= 2;
  }
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &b[i].first, &b[i].second);
    b[i].first *= 2;
    b[i].second *= 2;
  }
  solve();
  return 0;
}
