How to do variadic deduction in std::function's parameters?

I try to implement a function f: (std::function -> int) which will pass 1s into input_functor with c++ variadic template.

Let input_functor be g.

For example:

• If g is std::function<int(int,int)>, then f return g(1, 1).
• If g is std::function<int(int,int,int)>, then f return g(1, 1, 1).
• If g is std::function<int(int,int,int,int)>, then f return g(1, 1, 1, 1).

#include <functional>
#include <iostream>

template <typename T, typename... Args>
int apply(std::function<int(T, Args...)> func) {
    auto tmp = [func](Args... args) {
        return func(1, args...);
    };
    return apply(tmp);
}

template <typename T>
int apply(std::function<int(T)> func) {
    return func(1);
}

int main() {
    std::function<int(int, int)> f = [](int a, int b) {
        return a + b;
    };
    std::cout << apply(f) << "\n";
    return 0;
}

The compiler (clang++) error msg is that it cannot match candidates.

main.cpp:9:12: error: no matching function for call to 'apply'
    return apply(tmp);
           ^~~~~
main.cpp:21:18: note: in instantiation of function template specialization 'apply<int, int>' requested here
    std::cout << apply(f) << "\n";
                 ^
main.cpp:5:5: note: candidate template ignored: could not match 'function<int (type-parameter-0-0, type-parameter-0-1...)>' against
      '(lambda at main.cpp:6:16)'
int apply(std::function<int(T, Args...)> func) {
    ^
main.cpp:13:5: note: candidate template ignored: could not match 'function<int (type-parameter-0-0)>' against '(lambda at main.cpp:6:16)'
int apply(std::function<int(T)> func) {
    ^
1 error generated.