Failed parameter pack deduction of template arguments in a template specialization

I'm trying to write a code attempts a deduction of two parameter packs in template and was wondering why the following code doesn't work

t.h (doesn't work)

#pragma once

#include <tuple>

namespace example
{

template<typename ... T>
struct Foo;

template<typename ... T1, typename ... T2>
struct Foo<std::tuple<T1...>, std::tuple<T2...>>
{
    std::tuple<T1...> t1;
    std::tuple<T2...> t2;

    Foo(const std::tuple<T1...>& x, const std::tuple<T2...>& y) : t1(x), t2(y)
    {
    }
};
}

t.cpp

#include "t.h"

int main()
{
  using namespace example;
  using T1 = std::tuple<int>;

  example::Foo f(T1{}, T1{});

  return 0;
}

When I try to compile the code above, the compiler gives me the following error:

error: no viable constructor or deduction guide for deduction of template arguments of 'Foo'
        example::Foo f(T1{}, T1{});

After playing around a bit, I found out I can make the code work by adding a user defined deduction guide as follows

t.h (works)

namespace example
{

template<typename ... T>
struct Foo;

template<typename ... T1, typename ... T2>
struct Foo<std::tuple<T1...>, std::tuple<T2...>>
{
    std::tuple<T1...> t1;
    std::tuple<T2...> t2;

    Foo(const std::tuple<T1...>& x, const std::tuple<T2...>& y) : t1(x), t2(y)
    {
    }
};

// After adding this deduction guide, things work
template<typename T1, typename T2>
Foo(const T1& t1, const T2& out) -> Foo<T1, T2>;
}

The Question

My question is twofold. First, why does the compiler need a deduction guide in this case in the first place, and second, how does it work? I'm still learning the newer language features of C++ 20/23 and I'm using clang++ compiler.

Thanks for reading!