const char array template vs char const* function overload

I have looked at all the answers for the suggested related questions that stack overflow presented as I wrote this question, I have googled around, and I have looked at STL's video on overload resolution to try and figure out what makes the following code select the char const* overload over the template, but I can't figure it out.

#include <cstdio>

void foo( int i, char const* c_str_1, char const* c_str_2 )
{
   printf( "foo( int i, char const* c_str_1, char const* c_str_2 )\n" );
}

template< std::size_t N, std::size_t M >
void foo( int i, char const (&c_str_1)[N], char const (&c_str_2)[M] )
{
   printf( "foo( int i, char const (&c_str_1)[N], char const (&c_str_2)[M] )\n" );
}


int main( int argc, char* argv[] )
{
   char const* c_strg_1 = "This is a c style string.";
   foo( 1, c_strg_1, "bla" );
   foo( 1, "bla", "bla" );
   return 0;
}

It probably is something obvious I am missing, but the compiler always selects the char const* overload. I would have thought the template is an exact match and the other one needs a "decay"/"conversion".

Anyhow, thanks to anyone with insight into this.