Value of -faligned-new in GCC

I had an issue similar to this one gcc over-aligned new support (alignas )so I added -faligned-new compiler flag.

This fixed a compiler warning. However, when the compiler is run with --verbose flag, I get this output https://pastebin.com/X2QZAtSb, in which the most important line is

COLLECT_GCC_OPTIONS='-o''main''-faligned-new=1''-v''-shared-libgcc''-mtune=generic''-march=x86-64'

GCC manual (https://gcc.gnu.org/onlinedocs/gcc-9.3.0/gcc/C_002b_002b-Dialect-Options.html#C_002b_002b-Dialect-Options) states that:

-faligned-new

Enable support for C++17 new of types that require morealignment than void* ::operator new(std::size_t) provides. A numeric argument such as -faligned-new=32 can be used to specify how much alignment (in bytes) is provided by that function, but few users will need to override the default of alignof(std::max_align_t).

What does -faligned-new=1 mean? If I understand correctly, it should be equal to 8 or 16 or something (default of alignof(std::max_align_t), https://en.cppreference.com/w/cpp/types/max_align_t), not 1.

To reproduce compile simple main.cpp:

#include <iostream>int main() {    std::cout << "hello world\n";    return 0;}

using g++ main.cpp -o main -faligned-new --verbose

I am compiling the program on GCC 9.3.0, on a Debian system, sizeof(max_align_t) is 32 on my system.