#include <iostream>
#include <cstdint>
typedef uint32_t coord_t;
coord_t n, r1, c1, r2, c2;
bool blue_win;
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cin >> n;
std::cin >> r1 >> c1;
std::cin >> r2 >> c2;
// Ensure Blue is on the left side.
if (c1 > c2) {
c1 = n - c1 + 1;
c2 = n - c2 + 1;
}
coord_t sum = c1 + c2;
if (r1 == r2) {
coord_t columns_taken1 = sum >> 1;
blue_win = columns_taken1 > (n >> 1);
}
else {
// Let's assume r1 == 1 && r2 == 2
// They go towards each other.
// The question is what to do near the meet point?
coord_t diff = c2 - c1;
if (diff == 0) {
blue_win = false;
}
else if (diff == 1) {
coord_t cells_taken1 = std::max(
diff + 1 + ((n - c2) << 1), // moving right
sum - 1 // moving down
);
blue_win = cells_taken1 > n;
}
else if (diff & 1) {
// Let's calculate how many cells Blue can take if Red moves
// to the left near the meet point:
coord_t cells_taken1 = std::max(
diff + 1 + ((n - c2) << 1), // moving right
sum - 1 // moving down
);
// Now let's calculate how many cells Red would take:
coord_t cells_taken2 = std::max(
(n - (sum >> 1) - 1) << 1, // moving up
(n << 1) - cells_taken1 // moving left
);
blue_win = cells_taken2 < n;
}
else {
// Let's calculate how many cells Red can take if Blue moves
// to the right near the meet point:
coord_t cells_taken2 = std::max(
((c1 - 1) << 1) + (diff | 1), // moving left
(n << 1) - sum // moving up
);
// Now let's calculate how many cells Blue would take:
coord_t cells_taken1 = std::max(
sum - 2, // moving down
(n << 1) - cells_taken2 // moving right
);
blue_win = cells_taken1 > n;
}
}
if (blue_win)
std::cout << "Blue";
else
std::cout << "Red";
return 0;
}