MIPS has memory alignment restriction. If we'd store a 32-bit integer (i.e. 4 bytes) in memory locations and the current memory status as follows (shaded ones indicating memory occupied) how many different ways/locations you could store that integer. Note: the memory is byte addressed. Mem addr 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Mem content